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    Lagrange Multiplier Help

    Nice! Thanks a lot for the help! Problem resolved. :D
  2. L

    Lagrange Multiplier Help

    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)...
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    Lagrange Multiplier Help

    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...
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    Maximum Entropy Distribution Given Marginals

    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...
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    Lagrange Multiplier Help

    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...
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    Unbiased Estimator for Gamma Distribution Parameter

    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) =...
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    Counting Outcomes - Probability Question

    Doh! X = 0. Thats one more outcome. :( But my approach is correct?
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    Counting Outcomes - Probability Question

    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...
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    Probability Mass Function of { Z | Z < 1 }, Z given Z is less than 1

    hgfalling, thank you so much for all the probability help!
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    Probability Mass Function of { Z | Z < 1 }, Z given Z is less than 1

    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...
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    Density Function for Sums of Random Variables

    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...
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    Density Function for Sums of Random Variables

    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) =...
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    Support of Continuous Conditional Density Functions (Probability)

    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...
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    Support of Continuous Conditional Density Functions (Probability)

    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...
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    Jacobian of the linear transform Y = AX

    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!
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    Jacobian of the linear transform Y = AX

    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 =...
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    Is the gcd function symmetric?

    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) =...
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    Probability Density Function

    I asked the same question in this thread: https://www.physicsforums.com/showthread.php?t=398186 Hope it helps!
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    Probability density function of a random variable.

    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...
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    Probability density function of a random variable.

    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) =...
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    Deriving Joint Distribution Function from Joint Density (check my working pls!)

    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...
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    Deriving Joint Distribution Function from Joint Density (check my working pls!)

    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...
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    Probability density function of a random variable.

    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)
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    Probability density function of a random variable.

    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 -...
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    Deriving Joint Distribution Function from Joint Density (check my working pls!)

    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...
  26. L

    Sum of two continuous uniform random variables.

    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...
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    Sum of two continuous uniform random variables.

    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...
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    Deriving Joint Distribution Function from Joint Density (check my working pls!)

    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 {...
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    Indefinite Integration by exchange of variables - Problem 2

    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...
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    Indefinite Integration by exchange of variables

    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...
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