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Another CRV question (I tried)

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1. Homework Statement
The continuous random variable X has a uniform (rectangular) distribution on 0 ≤ x ≤ 1. Find the cumulative distributive function of X.

Three independent observations are made of X and the smallest of the three values is denoted by S.

i) Show that P(S ≤ x) = 3x - 3x² + x³ for 0 ≤ x ≤ 1
ii) Write down (or find) the value of E(S), and show that Var(S) = 3/80

The median of the three values is denoted by M. Find the probability density function of M.

2. Homework Equations
F(x) = ∫ f(x) dx
f(x) = 1 / (b-a) for a < x < b
Uniform distribution - E(X) = 1/2 (a + b)
Var (X) = 1/12 (b-a)²

3. The Attempt at a Solution
The first part is easy.

since f(x) = 1, F(X) = x for 0 ≤ x ≤ 1

F(X) =
{ 0, x < 0
{ x, 0 ≤ x ≤ 1
{ 1, x > 1

It is the part (i) and (ii) I'm stuck with.

Firstly, they said 3 independent observations and just the smallest value, which isn't known. I still don't understand why independent observations come into play here. Are this independent observations any number in the series or something? How do we know S from the other two values and determine M?
 
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Answers and Replies

tiny-tim
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Firstly, they said 3 independent observations and just the smallest value, which isn't known. I still don't understand why independent observations come into play here. Are this independent observations any number in the series or something? How do we know S from the other two values and determine M?
Hi PuzzledMe! :smile:

It's like drawing 3 cards from a suit of 13 cards … except you're drawing from a huge number of cards.

"independent observations" means that each observation is independent of the others, in the same sense as the second card will (usually) be independent of the first, and the third wil be independent of both of them!

If you like, pretend there's a million-and-one cards, numbered from 0 to a million.

You draw three … what is the chance that the lowest card is lower than card number x?

(Hint: sometimes it's easier to calculate the probability of the opposite.)

Does that help? :smile:
 
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Hi PuzzledMe! :smile:

It's like drawing 3 cards from a suit of 13 cards … except you're drawing from a huge number of cards.

"independent observations" means that each observation is independent of the others, in the same sense as the second card will (usually) be independent of the first, and the third wil be independent of both of them!

If you like, pretend there's a million-and-one cards, numbered from 0 to a million.

You draw three … what is the chance that the lowest card is lower than card number x?

(Hint: sometimes it's easier to calculate the probability of the opposite.)

Does that help? :smile:
I'm getting the point of the independent observations and thinking about it...
I think it is (x/million) chance but something tells me it may not be right.
 
tiny-tim
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… sorry … !

I'm getting the point of the independent observations and thinking about it...
I think it is (x/million) chance but something tells me it may not be right.
oh, sorry, I should have used some letter other than x … that's what's confusing! :redface:

I should have said "the card numbered y is at position x = y/100000. You draw three … what is the chance that the lowest card is lower than card number y?"
 
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If I'm not wrong on intepreting the definition of independent observation from what you've said, it means that for each of the three observations each result does not affect the consequence of the other results...

So the chance should be (y-1)/100000?
Or [x - (1/100000)]?

It's like if y is 2 then the only possible card that's lower is 1 and so its like one in 100000...
 
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tiny-tim
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Steady!

The probability that one card is lower than card number y is y/1000000.

But x = y/1000000.

So the probability that one observation is lower than x is y/1000000 … in other words, it's x.

And, contrarily, the probability that one observation is higher than x is 1 - x.

So what is the probability that the lowest of 3 observations is lower than x? :smile:
 
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Did you mean observation lower or equal to x? Or I'm on the wrong track here? Sorry not quite following you but I'm trying to...

P(X ≤ x) = x
P(X > x) = (1-x)

Stuck, need to sleep soon so I'll reply when I've thought it through...
Thanks tim.
 
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Steady!

The probability that one card is lower than card number y is y/1000000.

But x = y/1000000.

So the probability that one observation is lower than x is y/1000000 … in other words, it's x.

And, contrarily, the probability that one observation is higher than x is 1 - x.

So what is the probability that the lowest of 3 observations is lower than x? :smile:
So if say there is observation A, B and C (lowest to highest)...
I was thinking, will B and C affect observation A? First thought is no but I'm a little hesistant before I'm clear on this...
Could it be x? Or something that somehow leads to 3x - 3x² + x³?

Still thinking through...
 
tiny-tim
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I was thinking, will B and C affect observation A?
No … the question specifically says "Three independent observations …".

Hint: forget the word "observations". Just treat this as an ordinary pack-of-cards probability question.

If we say that Q is the statement "A < x", R is "B < x", and S is "C < x", with Q R and S independent, what is the probability of "Q or R or S"? :smile:
 
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No … the question specifically says "Three independent observations …".

Hint: forget the word "observations". Just treat this as an ordinary pack-of-cards probability question.

If we say that Q is the statement "A < x", R is "B < x", and S is "C < x", with Q R and S independent, what is the probability of "Q or R or S"? :smile:
Probability is x?

(Looking at this question makes me think I'm missing out some very important concept for this chapter...any idea what I should be reading up on to at least have a clue? I did fairly well for the statistics paper last year but I don't think it was the best I can do.)
 
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tiny-tim
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Hi PuzzledMe! :smile:

You really should be familiar with this trick:

P (Q or R or S)

= 1 - P(not-Q and not-R and not-S)

= 1 - P(not-Q)P(not-R)P(not-S).
Looking at this question makes me think I'm missing out some very important concept for this chapter...any idea what I should be reading up on to at least have a clue?
Sorry … I don't know about the books.

But probability is mostly common-sense.

You just have to think clearly … in this case, you have to ask yourself "what is a more convenient way of saying the lowest is lower than x?" (to whic the answer is "it is not the case that they are all higher than x.")

Keep practising, and you'll soon get the hang of it! :smile:

(btw, in this case you could also have guessed the answer by noticing that 3x - 3x² + x³ is very nearly something-cubed!)
 
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Hi PuzzledMe! :smile:

You really should be familiar with this trick:

P (Q or R or S)

= 1 - P(not-Q and not-R and not-S)

= 1 - P(not-Q)P(not-R)P(not-S).


Sorry … I don't know about the books.

But probability is mostly common-sense.

You just have to think clearly … in this case, you have to ask yourself "what is a more convenient way of saying the lowest is lower than x?" (to whic the answer is "it is not the case that they are all higher than x.")

Keep practising, and you'll soon get the hang of it! :smile:

(btw, in this case you could also have guessed the answer by noticing that 3x - 3x² + x³ is very nearly something-cubed!)
Thanks tim, somewhat worked things out to show part (i) of question. Part (ii) shouldn't be a problem, so now I have to think about the last one with M and S. I'm thinking that if S can now be worked out, I can try the reverse and then lead to M. Will see if it works here...

It looks like I really need to work out and understand probabilty doubly hard after finishing with this CRV chapter =(
 
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I have worked out (ii), but I'm stuck on the last part as I cannot visualise the median's relation to any of the stuff that I've worked out so far.

Steps for (ii) to show that I've been thinking:

F(S) = x³ - 3x² + 3x
f(S) = 3x² - 6x + 3

Mean of S
= E(S)
= [tex]\int^{\infty}_{-\infty} xf(x) dx[/tex]
= [tex]\int^{1}_{0} x(3x² - 6x + 3) dx[/tex]
= [tex]\int^{1}_{0} (3x³ - 6x² + 3x) dx[/tex]
= [tex][3x^{4} - 2x³ + \frac{3}{2} x²]^{1}_{0}[/tex]
= [3/4]

E(S²)
= [tex]\int^{\infty}_{-\infty} x²f(x) dx[/tex]
= [tex]\int^{1}_{0} x²(3x² - 6x + 3) dx[/tex]
= [tex]\int^{1}_{0} (3x^{4} - 6x³ + 3x²) dx[/tex]
= [tex][\frac{3}{5}^{5} - \frac{3}{2} x^{4} + x³]^{1}_{0}[/tex]
= [1/10]

Var (S)
= E(S²) - [E(S)]²
= 1/10 - 1/16
= 3/80
 
tiny-tim
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Hi PuzzledMe! :smile:

erm … you can't use unicode ² and ³ in LaTeX! :rolleyes:

I think you meant:
Mean of S
= E(S)
= [tex]\int^{\infty}_{-\infty} xf(x) dx[/tex]
= [tex]\int^{1}_{0} x(3x^2 - 6x + 3) dx[/tex]
= [tex]\int^{1}_{0} (3x^3 - 6x^2 + 3x) dx[/tex]
= [tex][3x^{4} - 2x^3 + \frac{3}{2} x^2]^{1}_{0}[/tex]
= [3/4]

E(S²)
= [tex]\int^{\infty}_{-\infty} x^2f(x) dx[/tex]
= [tex]\int^{1}_{0} x^2(3x^2 - 6x + 3) dx[/tex]
= [tex]\int^{1}_{0} (3x^{4} - 6x^3 + 3x^2) dx[/tex]
= [tex][\frac{3}{5} x^{5} - \frac{3}{2} x^{4} + x^3]^{1}_{0}[/tex]
= [1/10]
Fine … except the 3x^4 in E(S) should be (3/4)x^4. :smile:

:confused: what's the median … do they mean the middle of the three … no, I'm not testing you, I actually don't know … ?
 
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Opps my bad, first time using Latex.

It's something to do with this part of the question...

The median of the three values is denoted by M. Find the probability density function of M.
I read up on the CRV chapter notes and what they said about median is something along this...

Wikipedia said:
In probability theory and statistics, a median is described as the number separating the higher half of a sample, a population, or a probability distribution, from the lower half.
So it should be middle of the lowest value and highest value...
I don't know where to begin with here, but I'll continue to think on it.
 
tiny-tim
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So it should be middle of the lowest value and highest value...
I don't know where to begin with here, but I'll continue to think on it.
No … for an odd number (in this case, 3), wikipedia says the median is the middle one:
wikipedia said:
The median of a finite list of numbers can be found by arranging all the observations from lowest value to highest value and picking the middle one.
So you want the probability that the middle one is less than x;

which is the sum of the probabilities of {A and B < x, C > x} and {B and C < x, A > x} and {C and A < x, B > x};

which is … ? :smile:
 
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No … for an odd number (in this case, 3), wikipedia says the median is the middle one:


So you want the probability that the middle one is less than x;

which is the sum of the probabilities of {A and B < x, C > x} and {B and C < x, A > x} and {C and A < x, B > x};

which is … ? :smile:
I'll think about it and work this out tomorrow, late now...thanks tim, as always.

I think I'm beginning to see the bigger picture that there are a limited number of combinations and also a sort of "fixed" probability rate...

Mulling on it. :)
 
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Hi, sorry I know I asked this about 10 days plus ago but I felt I should come back to it because on checking on my workings with the answers, I'm still puzzled.

Answer says 6x - 6x², 0 ≤ x ≤ 1
My workings gave me 6x - 9x², 0 ≤ x ≤ 1
Am I doing it somewhere wrong?

My workings
P(M < x)
= P ( S ≤ x , T ≤ x, U > x) + P ( T ≤ x , U ≤ x, S > x) + P ( S ≤ x , U ≤ x, T > x)
= 3 (x) (x) (1 - x)
= 3 (x² - x³)
= 3x² - 3x³

p.d.f. of M = d/dx (3x² - 3x³) = 6x - 9x²

probability density function of M = m(x) = 6x - 9x², 0 ≤ x ≤ 1

any assistance will be really appreciated, I want to learn what went wrong and not make the same mistake again if any, thanks...
 
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tiny-tim
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oops!

Hi PuzzledMe! :smile:

ah … we started with the wrong formula …
So you want the probability that the middle one is less than x;

which is the sum of the probabilities of {A and B < x, C > x} and {B and C < x, A > x} and {C and A < x, B > x}
… we left out the probability of {A and B and C < x}.

Sorry! :redface:

Your workings were correct, so if you add that in, it should give the right answer! :smile:
 
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Thanks so much tim! It did =)
 

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