No, I meant to ask you about the problem on page 10. Your original reply solved my original problem.
To make it more precise:
Ignore everything that I have asked so far. The new problem is...
max - \Sigmax,y P(x,y) log P(x,y), subject to \Sigmay P(x,y) = r(x) for all x, and \Sigmax P(x,y)...
Yes thanks!
There is something else I don't get...
Page 10 of this lecture notes: http://pillowlab.cps.utexas.edu/teaching/CompNeuro10/slides/slides16_EntropyMethods.pdf [Broken]
Says we should get the solution as P(x,y) = P(x) P(y). I have no idea how to get this.
Furthermore, they seem to...
Hi all,
I'm a pure mathematician (Graph Theory) who has to go through a Physics paper, and I am having trouble getting through a part of it. Maybe you guys can point me in the right direction:
Let P(x,y) be a joint distribution function.
Let H = - \Sigmax,y P(x,y) log P(x,y), which is...
Homework Statement
L = - \Sigma x,y (P(x,y) log P(x,y)) + \lambda \Sigmay (P(x,y) - q(x))
This is the Lagrangian. I need to maximize the first term in the sum with respect to P(x,y), subject to the constraint in the second term.
The first term is a sum over all possible values of x,y...
Homework Statement
I need to show that c / (sample mean X) is unbiased for b, for some c.
Where {X} are iid Gamma (a,b).
Homework Equations
N.A.
The Attempt at a Solution
I know how to show that (sample mean X) / a is unbiased for 1/b :
1) E(sample mean X / a) =...
Homework Statement
Z plays a game where independent flips of a coin are recorded until two heads in succession are encountered.
Z wins if 2 heads in succession occurs.
Z loses if after 5 flips, we have not encounter two heads in succession.
1) What is the probability that Z wins the game...
Probability Mass Function of { Z | Z < 1 }, "Z given Z is less than 1"
Homework Statement
Given Z = X + Y.
Find the probability density function of Z|Z < 1.
Homework Equations
N.A.
The Attempt at a Solution
f(z) = P(Z=z|Z<1) = P(Z=z AND Z < 1) / P(Z < 1).
I thought the...
i came up with an answer on my own, is the process correct? i looked at the two arguments of f(x, z-x) and try to deduce which bounds are significant.
1) argument "x"
0 < x < z - x.
so x/2 < x < z/2. the z/2 is significant across all z in (0,2) since for z < 2, z/2 < 1. x < z/2 < 1, so...
Homework Statement
Given the joint density, f(x,y), derive the probability density function for Z = X + Y and V = Y - X.
Homework Equations
f(x,y) = 2 for 0 < x < y < 1
f(x,y) = 0 otherwise.
The Attempt at a Solution
For Z = X + Y, I can derive the fact that,
f_Z(z) =...
I just realized my mistake :
x is fixed since we are conditioning Y on a particular value of X = x. So the 0 < x < 1 is irrelevant. Or rather, the conditional probability density for a given X = x is a function of y only. So only the value of y matters in defining the support.
I feel like...
f(x,y) = x + y is the joint probability density function for continuous random variables X and Y. The support of this function is {0 < x < 1, 0 < y < 1}, which means it takes positive values over this region and zero elsewhere.
g(x) = x + (1/2) is the probability density function of X...
Thanks guys. I wrote Ax, for a constant martix A, as a linear combination of its columns, then deduce that each of the gi(X) is a linear combination of the entries in the ith row of A. Then the jacobian is the determinant of A transposed, which is equal to the determinant of A!
Homework Statement
Y = AX = g(X)
Where X,Y are elements of R^n and A is a nxn matrix.
What is the Jacobian of this transformation, Jg(x)?
Homework Equations
N.A.
The Attempt at a Solution
Well, I know what to do in the non-matrix case. For example...
U = g(x,y)
V =...
My input is probably trivial but I thought it would be nice to prove it formally.
We need : gcd(x,y) = gcd(y,x)
Suppose gcd(x,y) = d. Then x = nd and y = md for some integers n,m where gcd(n,m) = 1.
So gcd(y,x) = gcd (md,nd) = d because gcd(m,n) =1.
We have shown for all x,y that gcd(x,y) =...
You're welcomed man! Getting stuck at maths sucks! I got stuck for days recently and this forum helped me out, so I know how valuable maths help can be!
W = V^2
1) V has a continuous uniform density function over (-a,a) for a>0. Let g(v) be the density function for V. g(v) = 1/2a. The formula...
You are right. The answer is correct. Sorry about the confusion!
Cumulative Dist. = F(y) = 1 - G(-sqrt Y) + G(sqrt Y)
Density = f(y) = d/dy [F(y)] = (1/2sqrt(y))[ g(-sqrt Y) + g(sqrt Y)]
where g is the density of X.
g(-sqrt Y) = 0 since X is a positive random variable.
So, f(y) =...
OMG terrible terrible mistake there. Not sure what I was thinking at the time of posting!
Suppose to be:
F(x,y) = \int_{-\infty}^y \int_{-\infty}^x \! f(u,v) \, du dv
F(x,y) = \int_{0}^y \int_{0}^x \! f(u,v) \, du dv
But the 2nd expression for F(x,y) only holds for the region...
Oh...I see. Thanks a lot man, you saved my *** several times on questions related to probability! Setting up the integral...?
f(x,y) is defined for {0<x<1,0<y<1}
by definition,
F(x,y) = \int_{-\infty}^y \int_{-\infty}^x \! f(u,v) \, du dv
F(x,y) = \int_{0}^y \int_{0}^x \! f(x,y) \, dx dy...
Have you looked at my suggested method? After you differentiate the cumulative distribution function F(y) to get the probability density f(y), the answer is different from yours. Maybe there is a printing error. (happens all the time with textbooks)
My course teaches us to derive from basic definitions. Not sure if its the best way but its the only way I know. :p
(Cumulative) Distribution Function of Y
= F(y)
= P(Y < or = y)
= P(X^2 < or = y)
= P( -sqrt y < or = X < or = sqrt y)
= [ 1 - P(x < -sqrt y) ] + P(x < or = sqrt y)
= 1 -...
Oops, sorry! I forgot to adjust for notation differences. In my course, they call the joint cumulative distribution function the distribution function and etc. But YES, what you laid out was precisely the question I require assistance on.
I think I am fine with 90% of the working required for...
This logic seems to fail when considering X ~ exponential (L) and Y ~ exponential (L). Because X and Y have the exponential distribution, they take values in [0,infinity).
Z = X + Y, so z can take values in [0,infinity) too.
1) So should I replace the limits of integration from...
Please verify my logic in this attempt at a solution! I know my answer is correct but I want to know if my logical process is sound. :p
First, I need to determine where the "peak" is. For this I sketched the density of Z by considering values of z = 1,2,3,4,5. The peak appears to be at 3. So...
Making it shorter because I think the length is the reason why there are no replies...
f(x,y) = x + y, for 0<x<1 and 0<y<1
f(x,y) = 0, otherwise
f = joint density. we want the joint distribution function, F.
I only require asistance in verifying my logic for the expression for F over {...
Adding on to the other poster...
I prefer not to suppress the denominator.
So, u = 1 + 1/x
du/dx = (-1/x^2) dx/dx, since dx/dx = 1 you can multiply the rhs by it.
So, (-x^2) du/dx = dx/dx.
The dx in your integral can be seen as dx/dx. Substitute it with this value and we are done...
There appears to be many notations and definitions of "dy" where y is a dummy variable. Some authors chose to define dy, while others leave it at "no meaning". I came across one interpretation that might help you...
du/dx = 4 = 4 dx/dx
So, (1/4) du/dx = dx/dx
By suppressing the...