MHB Unsolved statistics questions from other sites, part II

[chisigma]

Re: Unsolved statistic questions from other sites, part II

... and you don't need a computer to evaluate it, tables are quite adequate...

CB

Unfortunately the personal experience of more that thirty five years in the area of telecommunications doesn't agree with this point of view. In the 'Bible' of Abramowitz and Stegun...

Abramowitz and Stegun: Handbook of Mathematical Functions

... the table of the normalized integral...$\displaystyle erf(x)= \frac{1}{\sqrt{2\ \pi}}\ \int_{- \infty}^{x} e^{- \frac{t^{2}}{2}}\ dt$ (1)

... arrives till to x=5 and supplies the value $\text{erf} (x) \sim .9999997133 \implies \text{erfc} (x) \sim 2.867 10^{-7}$. Well!... in digital transmission a standard bit error rate not greater that $10^{-6}$ is required, and that means that, in order to have necessary 'system margin', a target of bit error rate of $10^{-8} - 10^{-9}$ is often required... and even less in the case of optical fibre link...

At this point it is clear that the use of tables was for me not adequate, so that a lot of years ago I composed, with 'patient' application of the Simpson rule, the following 'little but accurate table' of the function $\log_{10} \text{erfc} (x)$, where 'erfc(x)' is defined as ...

$\displaystyle \text{erfc} (x) = 1 - \frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{- t^{2}}\ dt$ (2) View attachment 505

May be that, sooner or later, in a dedicate post, I will better explain the 'little accurate table' and indicate an easy way to transform it in a 'little computer program'... Kind regards $\chi$ $\sigma$

 

[CaptainBlack]

Re: Unsolved statistic questions from other sites, part II

Unfortunately the personal experience of more that thirty five years in the area of telecommunications doesn't agree with this point of view. In the 'Bible' of Abramowitz and Stegun...

Abramowitz and Stegun: Handbook of Mathematical Functions

... the table of the normalized integral...$\displaystyle erf(x)= \frac{1}{\sqrt{2\ \pi}}\ \int_{- \infty}^{x} e^{- \frac{t^{2}}{2}}\ dt$ (1)

... arrives till to x=5 and supplies the value $\text{erf} (x) \sim .9999997133 \implies \text{erfc} (x) \sim 2.867 10^{-7}$. Well!... in digital transmission a standard bit error rate not greater that $10^{-6}$ is required, and that means that, in order to have necessary 'system margin', a target of bit error rate of $10^{-8} - 10^{-9}$ is often required... and even less in the case of optical fibre link...

At this point it is clear that the use of tables was for me not adequate, so that a lot of years ago I composed, with 'patient' application of the Simpson rule, the following 'little but accurate table' of the function $\log_{10} \text{erfc} (x)$, where 'erfc(x)' is defined as ...

$\displaystyle \text{erfc} (x) = 1 - \frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{- t^{2}}\ dt$ (2) https://www.physicsforums.com/attachments/505

May be that, sooner or later, in a dedicate post, I will better explain the 'little accurate table' and indicate an easy way to transform it in a 'little computer program'... Kind regards $\chi$ $\sigma$

Then get a better table, mine goes to \(z=9.5\) with a tail probability of \(\sim 10^{-21}\), Also A&S give pretty good asymtotic representations for the extreme tails of the normal distribution (26.2.12 and following sections).

Also the suggestion of using a normal calculator may be less than useless to a student who will meet such a problem where they do not have access to calculation aides but may have an exam handbook with a table.

CB

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Then get a better table, mine goes to \(z=9.5\) with a tail probability of \(\sim 10^{-21}\)...

That's not a very difficuly task if we use the formula in...

Erfc -- from Wolfram MathWorld

$\displaystyle \frac{2}{\sqrt{\pi}}\ \frac{e^{- x^{2}}}{x + \sqrt{x^{2}+2}} < \text{erfc} (x) \le \frac{2}{\sqrt{\pi}}\ \frac{e^{- x^{2}}}{x + \sqrt{x^{2}+\frac{4}{\pi}}}$ (1)

... which gives an 'upper bound' and a 'lower bound' of the function. In the figure...

View attachment 506
... only the 'upper bound' is shown because the 'lower bound' in logaritmic scale is hard to be dinstinct from it. Pf course the only limitation in proceeding is the size of the diagram. It seems that the agreement with my old computation is good enough...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 12 12 2012 [the 'magic date' of the Maya's calendar!...] on www.mathhelpforum.com by the user asilvester635 and not yet solved…

While he was a prisoner of war during World War II, John Kerrich tossed a coin 10,000 times. He got 5067 heads. If the coin is perfectly balanced, the probability of a head is 0.5. Is there reason to think that Kerrich's coin was not balanced?... To answer this question use a normal distribution to estimate the probability that tossing a balanced coin 10,000 times would give a count of heads at least this far from 5000, that is, at least 5067 heads or no more than 4933 heads…

The problem is very similar to what treated in...

http://www.mathhelpboards.com/f23/unsolved-statistic-questions-other-sites-part-ii-1566/index3.html... and the requested probability is...

$\displaystyle P \sim \text {erfc} (\frac {5067.5 - \mu}{\sigma\ \sqrt{2}})$ (1)

... where $\mu = 10000\ p = 5000$ and $\sigma= \sqrt{10000\ p\ (1-p)}= 50$ . For $x = .9546$ 'MonsterWolfram' supplies $\displaystyle \text{erfc} (x) \sim .177$, so that the Kerric's coin seems to be a little unbalanced toward head. The scope of this post however is to verify the possibility to use the approximate value of the erfc(*) described in...

http://www.mathhelpboards.com/f23/unsolved-statistic-questions-other-sites-part-ii-1566/index4.html#post12076

... by the formula...

$\displaystyle \frac{2}{\sqrt{\pi}}\ \frac{e^{- x^{2}}}{x + \sqrt{x^{2}+ 2}} < \text {erfc} (x) \le \frac{2}{\sqrt{\pi}}\ \frac{e^{- x^{2}}}{x + \sqrt{x^{2}+ \frac{4}{\pi}}}$ (2)

Using a normal handset calculator for $x = .9546$ we find...

$ \displaystyle .170483 < \text{erfc(.9546)} < .186478$

... and taking the aritmetic mean $\text{erfc(.9546)} \sim .1784$, a result 'good enough' obtained without using tables...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted on 12 15 2012 on www.artofproblemsolving.com by the member BlackMax and not yet solved...

Three points are uniformly and independently chosen inside a given circle. What is the probability that their circumcircle lies entirely within the given circle?... a C++ program suggests that the answer is most likely to be .4 ...Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted on 12 15 2012 on www.artofproblemsolving.com by the member BlackMax and not yet solved...

Three points are uniformly and independently chosen inside a given circle. What is the probability that their circumcircle lies entirely within the given circle?... a C++ program suggests that the answer is most likely to be .4 ...

Clearly a 'direct' attack to this problem is a little unconfortable so that I'll try and 'indirect' attack. Let's suppose that the circle is the unit circle and that a 'circumcentre point' can be represented by the distance r from the point [0,0], as il the figure... View attachment 549

If we fix the circumcentre, then the mesure of the set of possible 'random points' is the area of the 'red circle' in the figure, so that the requested probability is given by the 'simple' computation...

$\displaystyle P = \int_{0}^{1} (1-r)^{2}\ d r = \frac{1}{3}$ (1)

Honestly however I'm not 'fully certain' of my solution and some suggestion and/or comments from MHB members is wellcome... Kind regards chi sigma

 

[Nono713]

Re: Unsolved statistic questions from other sites, part II

Chisigma, I don't believe your answer is correct. You seem to be assuming the circumcentre has the same distribution as the three randomly selected points, it clearly doesn't follow the same distribution. Consider what happens when the three random points are almost colinear, for instance (the circumcircle grows much larger than the unit circle)

I wrote a little Python 3.2 script to try and calculate the probability of the circumcircle of three random points in the unit circle being fully contained in the unit circle:

from math import cos, sin, sqrt, pi
from random import random

''' Generates 'n' random points uniformily distributed in the unit circle. '''
def RandomPoints(n):
    points = []

    for t in range(n):

        # Uniform polar distribution
        theta = random() * 2 * pi
        radius = random()
        
        # Convert to cartesian
        x = radius * cos(theta)
        y = radius * sin(theta)

        points.append({'x': x, 'y': y})

    return points

''' Returns the circumcircle of three points a, b, c. '''
def Circumcircle(a, b, c):

    # Translate such that vertex a is at the origin
    u = {'x': b['x'] - a['x'], 'y': b['y'] - a['y']}
    v = {'x': c['x'] - a['x'], 'y': c['y'] - a['y']}

    # Precompute a few values
    d = 2 * (u['x'] * v['y'] - u['y'] * v['x'])
    e = pow(u['x'], 2) + pow(u['y'], 2)
    f = pow(v['x'], 2) + pow(v['y'], 2)

    # Compute the circumcentre's translated coordinates
    centre = {'x': (v['y'] * e - u['y'] * f) / d,
              'y': (u['x'] * f - v['x'] * e) / d}

    # Compute the circumcircle's radius (note: translated)
    radius = sqrt(pow(centre['x'], 2) + pow(centre['y'], 2))

    # Translate the circumcentre back into the original space
    centre = {'x': centre['x'] + a['x'], 'y': centre['y'] + a['y']}
    
    return {'centre': centre, 'radius': radius}

''' Verifies if the given circle is fully contained by the unit circle. '''
def InUnitCircle(circle):
    
    # Check if the given circumcircle is completely inside the unit circle
    d = sqrt(pow(circle['centre']['x'], 2) + pow(circle['centre']['y'], 2))

    if d > 1 + circle['radius']:
        return False # do not intersect

    if d <= abs(1 - circle['radius']):
        return True # contained

    return False # overlap

def Experiment(trials):
    passed = 0

    for t in range(trials):

        p = RandomPoints(3)
        c = Circumcircle(p[0], p[1], p[2])

        if InUnitCircle(c):
            passed += 1

    return passed / trials

# Experiment script
for t in range(2, 10):
	prob = Experiment(pow(10, t))
	print("After " + str(pow(10, t)) + " trials, P ~ " + str(prob))

These are the probabilities I measured:

ray@ray ~/circumcentre $ python3 circumcentre.py
After 100 trials, P = 0.6
After 1000 trials, P = 0.446
After 10000 trials, P = 0.4651
After 100000 trials, P = 0.46801
After 1000000 trials, P = 0.465733
After 10000000 trials, P = 0.4654937
After 100000000 trials, P = 0.46576991
...

However, it is possible I made an implementation mistake, please let me know if you see one!

I don't know how to attack the problem mathematically, though - deriving the circumcentre's distribution seems quite infeasible (there must be some kind of trick) so I thought I'd go numerical for this one (Smoking)

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Thanks to Bacterius for the remarkable work!...

An insidious trap in this problem is connected to the definition of 'random point inside the unit circle'. Some months ago that has been topic of discussion in...

http://www.mathhelpboards.com/f23/unsolved-statistics-questions-other-sites-932/index2.html#post5428
http://www.mathhelpboards.com/f23/unsolved-statistics-questions-other-sites-932/index2.html#post5430
http://www.mathhelpboards.com/f23/unsolved-statistics-questions-other-sites-932/index2.html#post5431
http://www.mathhelpboards.com/f23/unsolved-statistics-questions-other-sites-932/index2.html#post5433
http://www.mathhelpboards.com/f23/unsolved-statistics-questions-other-sites-932/index2.html#post5434
http://www.mathhelpboards.com/f23/unsolved-statistics-questions-other-sites-932/index2.html#post5434

... and the conclusion was that the correct definition is the following: a random point is uniformely distributed inside the unit circle if the probability to find it in an area A is proportional to A. The consequence is that, indicating the point as complex number $z= \rho\ e^{i\ \theta}$, in order to respect the definition above, $\rho$ is not uniformely distributed in [0,1] but has a p.d.f. of the type...

$f(x) = \begin{cases} 2\ x & \text{if } 0<x<1\\
0 & \text{otherwise } \end{cases}$ (1)

If You generate a great number of points with $\rho$ ditributed according to (1) You obtain an uniform distribution inside the unit circle whereas with $\rho$ uniformely distributed You obtain concentration ofr points around the center. I think that could be useful to repeat simulation using $\rho$ distributed as in (1)... Kind regards $\chi$ $\sigma$

 

[Nono713]

Re: Unsolved statistic questions from other sites, part II

Thanks to Bacterius for the remarkable work!...

An insidious trap in this problem is connected to the definition of 'random point inside the unit circle'. Some months ago that has been topic of discussion in...

http://www.mathhelpboards.com/f23/unsolved-statistics-questions-other-sites-932/index2.html#post5428
http://www.mathhelpboards.com/f23/unsolved-statistics-questions-other-sites-932/index2.html#post5430
http://www.mathhelpboards.com/f23/unsolved-statistics-questions-other-sites-932/index2.html#post5431
http://www.mathhelpboards.com/f23/unsolved-statistics-questions-other-sites-932/index2.html#post5433
http://www.mathhelpboards.com/f23/unsolved-statistics-questions-other-sites-932/index2.html#post5434
http://www.mathhelpboards.com/f23/unsolved-statistics-questions-other-sites-932/index2.html#post5434

... and the conclusion was that the correct definition is the following: a random point is uniformely distributed inside the unit circle if the probability to find it in an area A is proportional to A. The consequence is that, indicating the point as complex number $z= \rho\ e^{i\ \theta}$, in order to respect the definition above, $\rho$ is not uniformely distributed in [0,1] but has a p.d.f. of the type...

$f(x) = \begin{cases} 2\ x & \text{if } 0<x<1\\
0 & \text{otherwise } \end{cases}$ (1)

If You generate a great number of points with $\rho$ ditributed according to (1) You obtain an uniform distribution inside the unit circle whereas with $\rho$ uniformely distributed You obtain concentration ofr points around the center. I think that could be useful to repeat simulation using $\rho$ distributed as in (1)... Kind regards $\chi$ $\sigma$

Ooh! You are absolutely right, my random sampling is incorrect. Stupid me! (Fubar) Here is the corrected script with the proper PDF for the radius:

from math import cos, sin, sqrt, pi
from random import random

''' Generates 'n' random points uniformily distributed in the unit circle. '''
def RandomPoints(n):
    points = []

    for t in range(n):

        # Uniform polar distribution
        theta = random() * 2 * pi
        radius = sqrt(random())
        
        # Convert to cartesian
        x = radius * cos(theta)
        y = radius * sin(theta)

        points.append({'x': x, 'y': y})

    return points

''' Returns the circumcircle of three points a, b, c. '''
def Circumcircle(a, b, c):

    # Translate such that vertex a is at the origin
    u = {'x': b['x'] - a['x'], 'y': b['y'] - a['y']}
    v = {'x': c['x'] - a['x'], 'y': c['y'] - a['y']}

    # Precompute a few values
    d = 2 * (u['x'] * v['y'] - u['y'] * v['x'])
    e = pow(u['x'], 2) + pow(u['y'], 2)
    f = pow(v['x'], 2) + pow(v['y'], 2)

    # Compute the circumcentre's translated coordinates
    centre = {'x': (v['y'] * e - u['y'] * f) / d,
              'y': (u['x'] * f - v['x'] * e) / d}

    # Compute the circumcircle's radius (note: translated)
    radius = sqrt(pow(centre['x'], 2) + pow(centre['y'], 2))

    # Translate the circumcentre back into the original space
    centre = {'x': centre['x'] + a['x'], 'y': centre['y'] + a['y']}
    
    return {'centre': centre, 'radius': radius}

''' Verifies if the given circle is fully contained by the unit circle. '''
def InUnitCircle(circle):
    
    # Check if the given circumcircle is completely inside the unit circle
    d = sqrt(pow(circle['centre']['x'], 2) + pow(circle['centre']['y'], 2))

    if d > 1 + circle['radius']:
        return False # do not intersect

    if d <= abs(1 - circle['radius']):
        return True # contained

    return False # overlap

def Experiment(trials):
    passed = 0

    for t in range(trials):

        p = RandomPoints(3)
        c = Circumcircle(p[0], p[1], p[2])

        if InUnitCircle(c):
            passed += 1

    return passed / trials

# Experiment script
for t in range(2, 10):
	prob = Experiment(pow(10, t))
	print("After " + str(pow(10, t)) + " trials, P ~ " + str(prob))

And the results:

After 100 trials, P ~ 0.3
After 1000 trials, P ~ 0.406
After 10000 trials, P ~ 0.4056
After 100000 trials, P ~ 0.40125
After 1000000 trials, P ~ 0.400055

Essentially 0.4, as suggested by the original poster.

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 01 25 2013 on www.talkstats.com by the user zeion and not yet solved...

Assume that an application sends packets over a communication link with a capacity of C
bits per second, i.e. when the packet is i bits long, then it takes $\displaystyle \frac{i}{C}$ seconds to transmit
the entire packet. Assume that the length L (in bits) of a packet is given by a random
variable with a geometric distribution, i.e.

$\displaystyle P(L = i) = \mu (1 − \mu)^{i-1},\ i \ge 1$

(a) Find the expect length E[L] of a packet as a function of $\μ$

(b) Find the probability that it takes t seconds to transmit a packet for for t > 0

(c) Find the expected time to transmit a packet. How does the expected time change
when we double the capacity C? How does the expected time change when we double
the expected packet length E[L]?

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 01 25 2013 on www.talkstats.com by the user zeion and not yet solved...

Assume that an application sends packets over a communication link with a capacity of C
bits per second, i.e. when the packet is i bits long, then it takes $\displaystyle \frac{i}{C}$ seconds to transmit
the entire packet. Assume that the length L (in bits) of a packet is given by a random
variable with a geometric distribution, i.e.

$\displaystyle P(L = i) = \mu (1 − \mu)^{i-1},\ i \ge 1$

(a) Find the expect length E[L] of a packet as a function of $\mu$

(b) Find the probability that it takes t seconds to transmit a packet for for t > 0

(c) Find the expected time to transmit a packet. How does the expected time change
when we double the capacity C? How does the expected time change when we double
the expected packet length E[L]?

From the well known geometric series...

$\displaystyle \sum_{i=0}^{\infty} x^{i} = \frac{1}{1-x}$ (1)

... we derive...

$\displaystyle \sum_{i=1}^{\infty} i\ x^{i-1} = \frac{d}{d x} \frac{1}{1-x}= \frac{1}{(1-x)^{2}}$ (2)

... so that is...

$\displaystyle E\{L\} = \mu\ \sum_{i=1}^{\infty} i\ (1-\mu)^{i-1} = \frac{\mu}{\mu^{2}} = \frac{1}{\mu}$ (3)

That answers to a) and from it the answers to b) and c) follow immediately. Of course that is a fully elementary statistical problem, but You have to know that I have spent somewhat like 35 years of my life working in telecommunications area so that(Happy)...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 02 10 2013 on mathhelpforum.com by the user bullmoose97 and not yet solved...

Let X1, X2, X3, X4 be independent random variables with common density function:f(x) = e^-x, x>0 and 0, elsewhere. Let U = X1 + X2 + X3 + X4. Find the density function of U.

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 02 10 2013 on mathhelpforum.com by the user bullmoose97 and not yet solved...

Let X1, X2, X3, X4 be independent random variables with common density function:f(x) = e^-x, x>0 and 0, elsewhere. Let U = X1 + X2 + X3 + X4. Find the density function of U.

If f(x) is the p.d.f. of the X, then the p.d.f. of U is f(x)*f(x)*f(x)*f(x) where '*' means convolution. In our case is $\displaystyle f_{X}(x)=e^{-x}$ and because is $\displaystyle \mathcal{L} \{e^{-x}\}= \frac{1}{1+s}$ it will be...

$\displaystyle f_{U}(x) = \mathcal{L}^{-1} \{\frac{1}{(1+s)^{4}}\} = \frac{1}{6}\ x^{3}\ e^{-x}$ (1)

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted on 02 15 2013 on www.artofproblemsolving.com by the user mynamearzo and not yet solving...

Let $(X,Y)$ be a point chosen at random from the circle with centre $\displaystyle (\frac{1}{2},\frac{1}{2})$ and radius $\displaystyle \frac{1}{2}$. Show that X and Y are uniformly distributed over the unit interval. Are they independent?...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted on 02 15 2013 on www.artofproblemsolving.com by the user mynamearzo and not yet solving...

Let $(X,Y)$ be a point chosen at random from the circle with centre $\displaystyle (\frac{1}{2},\frac{1}{2})$ and radius $\displaystyle \frac{1}{2}$. Show that X and Y are uniformly distributed over the unit interval. Are they independent?...

Of course nothing changes if we translate the coordinates such that the circle in centered in (0,0) as in the figure...

https://www.physicsforums.com/attachments/625._xfImport

As explained in…

http://www.mathhelpboards.com/f23/unsolved-statistic-questions-other-sites-part-ii-1566/index4.html#post13740

… the correct definition is the following: a random point is uniformly distributed inside the unit circle if the probability to find it in an area A is proportional to A and in this case A is the colored area in the figure. With this definition is...

$\displaystyle P \{X \le x\} = 1- \frac{\cos ^{-1} 2x}{\pi} + \frac{\pi}{4}\ x\ \sqrt{\frac{1}{4}-x^{2}}$ (1)

The p.d.f. of the r.v. X is obtained deriving (1)...

$\displaystyle f_{X} (x)= \frac{\pi^{2}\ (1-2 x^{2}) + 4}{4\ \pi\ \sqrt{1 - x^{2}}}$ (2)

... so that X isn't uniformly distributed in $[- \frac{1}{2},\frac{1}{2}]$, even if the approximation is good, as demonstrated by the plot of the (1)...

https://www.physicsforums.com/attachments/624._xfImport
Of course critics and observation from other MHB fellows are wellcome!...By symmetry the $f_{y}(y)$ is quite identical to (2). But are X and Y independent?... the answers will be given in a next post... Kind regards $\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 02 16 2013 on www.artofproblemsolving.com by the user mynamearzo and not jet solved... Show that for the standard normal distribution, Mill's ratio [defined as $\displaystyle \frac{1-F(x)}{f(x)}$ where $F(x)$ is the distribution function and $f(x)$ is the density function] has upper and lower bounds given respectively by $\displaystyle \frac{1}{x}$ and $\displaystyle \frac{1}{x} - \frac{1}{x^{3}}$. Can you improve upon the bounds?...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 02 16 2013 on www.artofproblemsolving.com by the user mynamearzo and not jet solved...Show that for the standard normal distribution, Mill's ratio [defined as $\displaystyle \frac{1-F(x)}{f(x)}$ where $F(x)$ is the distribution function and $f(x)$ is the density function] has upper and lower bounds given respectively by $\displaystyle \frac{1}{x}$ and $\displaystyle \frac{1}{x} - \frac{1}{x^{3}}$. Can you improve upon the bounds?...

The first step is of course the precise definition of Mill's ratio...$\displaystyle \mathcal{m}(x)= \frac{1-F(x)}{f(x)} $ (1)

... and in the case of standard normal distribution is...

$\displaystyle f(x) = \frac{e^{- \frac{x^{2}}{2}}}{\sqrt{2 \pi}} \implies 1-F(x)= \frac{1}{\sqrt{2 \pi}}\ \int_{x}^{\infty} e^{- \frac{u^{2}}{2}}\ du$ (2)

... so that is...

$\displaystyle \mathcal{m}(x)= e^{\frac{x^{2}}{2}}\ \int_{x}^{\infty} e^{- \frac{u^{2}}{2}}\ du$ (3)

More than seventy years ago R. D. Gordon proved that...$\displaystyle \frac{x}{1+x^{2}} \le \mathcal{m}(x) \le \frac {1}{x}$ (4)

... and because is...

$\displaystyle \frac{x}{1+x^{2}} = \frac{1}{x}\ \frac{1}{1+\frac{1}{x^{2}}} \sim \frac{1}{x} - \frac{1}{x^{3}}$ (5)

... we obtain...

$\displaystyle \frac{1}{x} - \frac{1}{x^{3}} \le \mathcal{m}(x) \le \frac {1}{x}$ (6)

Some time after Z. W. Birnbaum improved the (4) proving that...

$\displaystyle \frac{\sqrt{4+x^{2}}- x}{2} \le \mathcal{m}(x) \le \frac {1}{x}$ (7)http://projecteuclid.org/DPubS/Repo...w=body&id=pdf_1&handle=euclid.aoms/1177731611

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 03 17 2013 on www.artofproblemsolving.com by the user gauss202 and not yet solved…

A standard 52-card deck of cards is shuffled, and you draw cards from the top until you see an ace. On average, how many cards will you draw?...Kind regards $\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 03 17 2013 on www.artofproblemsolving.com by the user gauss202 and not yet solved…

A standard 52-card deck of cards is shuffled, and you draw cards from the top until you see an ace. On average, how many cards will you draw?...

The probability that the game ends after one draw is $\displaystyle \frac{4}{52}$, after two draws is $\displaystyle \frac{4}{51} (1- \frac{4}{52})$ and after n draws ... $\displaystyle p_{n}= \frac{4}{52-n+1} (1-\frac{4}{52}) \cdot (1-\frac{4}{51}) ... (1-\frac{4}{52-n+2}) = 4\ \frac{48 \cdot 47 ... (48 -n+2)}{52 \cdot 51 ... (52-n+1)} $ (1)

... so that the expected value of n is...

$\displaystyle E \{n\} = \sum_{n=1}^{49} n\ p_{n}$ (2)

The search of a comfortable [if any...] way to valuate sum (2) will be performed in a successive post... any help will be of course well accepted! (Nod)...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted on 03 11 2013 on www.artofproblemsolving.com by the user hallohola and not yet solved...

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[TD="class: post-message"]Let random variables $\displaystyle X_{1} \ge X_{2} \ge ... \ge X_{n}$ be uniformly and independently distributed within [0,1]and let 0 < r < 1. Given that $\displaystyle X_{2} \ge r$ find the expected value of $X_{2}$

Kind regards

$\chi$ $\sigma$
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[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted on 03 11 2013 on [URL="http://www.artofproblemsolving.com"]www.artofproblemsolving.com[/URL] by the user hallohola and not yet solved...

Let random variables $\displaystyle X_{1} \ge X_{2} \ge … \ge X_{n}$ be independent uniformly distributed r.v. within [0,1] and let 0 < r < 1. Given that $\displaystyle X_{2} \ge r$ find the expected value of $X_{2}$...

The problem is solved 'step by step' and the first step is to find the expected value of the greatest of the r.v. $X_{1} \ge X_{2} \ge ... \ge X_{n}$ uniformly distributed in [0,1]. The probability that all the r.v. are in [0,x] is...

$\displaystyle P_{1} (x)= x^{n}$ (1)

... so that deriving (1) we obtain...

$\displaystyle f_{1} (x) = n\ x^{n-1}$ (2)

... and is...

$\displaystyle E \{X_{1}\}= \int_{0}^{1} n\ x^{n} dx = \frac{n}{n+1}$ (3)The second step is, given the same condition, to find the expected value of $X_{2}$. The probability that n-1 r.v. are in [0,x] and one r.v. is in [x,1] is...
$\displaystyle P_{2} = x^{n-1}$ (4)... so that we have...

$\displaystyle f_{2} (x) = (n-1)\ x^{n-2}$ (5)

$\displaystyle E \{X_{2}\}= \int_{0}^{1} (n-1)\ x^{n-1} dx = \frac{n-1}{n}$ (6)

The third step is to find the expected value of $X_{2}$ with the condition $X_{2} \ge r$ that is...

$\displaystyle E \{X_{2}\}_{X_{2} \ge r} = \frac{1}{1-r^{n-1}}\ \int_{r}^{1} (n-1)\ x^{n-1}\ dx = \frac{1-r^{n}}{1-r^{n-1}}\ \frac{n-1}{n}$ (7) Kind regards…

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 04 09 2013 on www.mathhelpforum.com by the user Kibbygirl and not jet solved...

According to a certain book, men have 14.334% chance of dying at age 90. A group of 90 year old men meet monthly to discuss philosophy and politics. They are discussing the possibility of some of their group dying this year. If 2 men are chosen at random from the group, what is the probability that both of the 2 men will die this year?...

Kind regards

$\chi$ $\sigma$​

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 04 09 2013 on www.mathhelpforum.com by the user Kibbygirl and not jet solved...

According to a certain book, men have 14.334% chance of dying at age 90. A group of 90 year old men meet monthly to discuss philosophy and politics. They are discussing the possibility of some of their group dying this year. If 2 men are chosen at random from the group, what is the probability that both of the 2 men will die this year?...

​

This problem has been choosen not because it is 'difficult' but because it is a good example of conditional probability. Let suppose that the cumulative distribution of life time $\lambda$ is, normalizing to 1 a time of 100 years, something like...

$\displaystyle P \{\lambda \le x\} = x^{\alpha}$ (1)

... where $\alpha$ is an unknown parameter that we can evalutate knowing that...

$\displaystyle P \{.9 \le \lambda \le 1\}= .14334$ (2)

From (2) we derive easily $\displaystyle \alpha= 1.4684...$ so that we can go to the second part of the problem. For semplicity sake let suppose that the men are both 90 and we want to evaluate the probability that one of them dies between 90 and 91 once he has lived till 90 that is...

$\displaystyle P\{ .9 \le \lambda \le .91\} = \frac{.91^{\alpha} - .9^{\alpha}}{.14334} = .09776...$ (3)

The probability that both men die this year is the square of (3)...

Kind regards

$\chi$ $\sigma$

​

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 10 07 2012 on www.scienzematematiche.it by the user pic [original in Italian...] and not yet solved...

Let be $X_{k}$ a sequence of independent r.v. uniformly distributed in [0,1]. How many variable in average do You need to be $\displaystyle X_{1} + X_{2} + ... + X_{n} > 1$?... Kind regards $\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 10 07 2012 on www.scienzematematiche.it by the user pic [original in Italian...] and not yet solved...

Let be $X_{k}$ a sequence of independent r.v. uniformly distributed in [0,1]. How many variable in average do You need to be $\displaystyle X_{1} + X_{2} + ... + X_{n} > 1$?...

The Laplace Transform of the p.d.f. of a single r.v. $X_{k}$ is...

$\displaystyle \mathcal{L} \{f_{1} (t)\} = \frac{1-e^{- s}}{s}$ (1)

... so that the Laplace Transform of the p.d.f. of the r.v. $\displaystyle X=\sum_{k=1}^{n} X_{k}$ is... $\displaystyle \mathcal{L} \{f_{n} (t)\} = \frac{1}{s^{n}}\ \sum_{k=0}^{n} (-1)^{k}\ \binom{n}{k}\ e^{- k s}$ (2)

... and from (2) ...

$\displaystyle f_{n} (t) = \frac{1}{(n-1)!}\ \sum_{k=0}^{n} (-1)^{k}\ \binom {n}{k}\ (t-k)^{n-1}\ \mathcal {U} (t-k)$ (3)

From (3) we derive the probability...

$\displaystyle P_{n} = P \{X < 1\} = \int_{0}^{1} \frac{t^{n-1}}{(n-1)!} \ dt = \frac{1}{n!}$ (4)

... so that is...

$\displaystyle E \{n\} = \sum_{n=0}^{\infty} n\ P_{n} = e$ (5)

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted on 04 03 2013 on www.artofproblemsolving.com by the user newsun and not yet solved ...

Number of men and women arriving at a shop in an hour are independently Poisson distributed with mean $\lambda$ and $\mu$ respectively and suppose that there are 3 people arrived during the first hour. What is the probability that...

a) no one arrived during the first 20 minutes...

b) one man and one woman arrived during the first 20 minutes...

c) the first customer is woman...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted on 04 03 2013 on www.artofproblemsolving.com by the user newsun and not yet solved ...

Number of men and women arriving at a shop in an hour are independently Poisson distributed with mean $\lambda$ and $\mu$ respectively and suppose that there are 3 people arrived during the first hour. What is the probability that...

a) no one arrived during the first 20 minutes...

b) one man and one woman arrived during the first 20 minutes...

c) the first customer is woman...

The distributions are both Poisson so that the probability that k men arrived in one hour is $\displaystyle P_{m} (k) = e^{- \lambda}\ \frac{\lambda^{k}}{k!}$ and the probability that k women arrived in one hour is $\displaystyle P_{w} (k) = e^{- \mu}\ \frac{\mu^{k}}{k!}$, and we use them to compute the probability $P_{3}$ that 3 people arrived in the first hour...$\displaystyle P_{3} = \sum_{k=0}^{3} e^{-\lambda}\ e^{- \mu} \frac{\lambda^{k}\ \mu^{3-k}}{k!\ (3-k)!} = \frac{e^{- (\lambda+ \mu)}}{3!} \sum_{k=0}^{3} \binom{3}{k} \lambda^{k}\ \mu^{3-k} = \frac{e^{- (\lambda+ \mu)}}{3!}\ (\lambda + \mu)^{3}$ (1)

All the requested probabilities are conditioned to the event that 3 people arrived in the first hour the probability of which is the (1) and precisely...a) the probability that no men and no women arrived in the first 20 minutes is...$\displaystyle P_{0 + 0} = \frac{e^{- \frac{\lambda + \mu}{3}}}{P_{3}}$ (2)

b) the probability that one man and one woman arrived in the first 20 minutes is...$\displaystyle P_{1 + 1} = \lambda\ \mu\ \frac{e^{- \frac{\lambda + \mu}{3}}}{3\ P_{3}}$ (3)

The question c) is a little more subtle and requires further considerations...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

The answer to question c) is the conditioned probability that the first was a woman and that means that the field of events is restricted to have a woman and she arrived first so that it is...

$\displaystyle P_{\text{woman first}} = \frac{e^{- (\lambda + \mu)}}{P_{3}}\ (\mu + \sum_{k=2}^{3} \frac{\mu^{k}\ \lambda^{3-k}}{k!\ (3-k)!})$ (1)

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted on 04 03 2013 on www.artofproblemsolving.com by the user newsun and not yet solved ...

The number of cars entering a car park on a given day is Poisson distributed with mean 100 and the parking fee is 2 dollars per hour with minimum fee of 1 dollar. The amount of time a car is in the car park is exponential distributed with mean 30 minutes. Find the mean and the variance of amount of money that the car park takes in on a given day...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted on 04 03 2013 on www.artofproblemsolving.com by the user newsun and not yet solved ...

The number of cars entering a car park on a given day is Poisson distributed with mean 100 and the parking fee is 2 dollars per hour with minimum fee of 1 dollar. The amount of time a car is in the car park is exponential distributed with mean 30 minutes. Find the mean and the variance of amount of money that the car park takes in on a given day...

Assuming as unity time 30 minutes the fee is constant (1 dollar) for 0<t<1 and linearly increasing with t for t>1 so that the expected amount of dollars for each car is...

$\displaystyle E\{d\} = 1 + \int_{1}^{\infty} (t-1)\ e^{-t}\ dt = 1 + \frac{1}{e} = 1.367879...$ (1)

The number of cars enetering the park on a given day is Poisson distributed with mean 100, so that the expect daily gain of the park is $\displaystyle 100\ (1+\frac{1}{e}) = 136.7879...\ \text{dollars}\ $...

Kind regards

$\chi$ $\sigma$