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Thread: Maximum of a set of independent normal random variables

  1. #16
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    Not gonna go into the derivation but I did some simulations and it appears to match what we want. Maybe when I have some free time I'll type up how I did it but essentially I said let X1 be the variable of interest and we want to know P(X1 > max{X2,...,Xn})

    I say that this probability is equal to:

    let f_i(x) denote the pdf of the ith random variable
    let F_i(x) denote the cdf of the ith random variable
    then
    P(X_1 >max\{X_2,\ldots,X_n\}) = \1 - \int_{-\infty}^\inftyF_1(x)f_2(x)F_3(x)\cdots F_n(x)\  +\ F_1(x)F_2(x)f_3(x)F_4(x)\cdots F_n(x) + \ldots + F_1(x)F_2(x)\cdots F_{n-1}(x)f_n(x) \  dx


    You can simulate this pretty easily by noticing that each term in the integral is just an expected value. So you could simulate the first part of the integral by generating a bunch of observations from f_2(x) and then apply F_1(x)F_3(x)\cdots F_n(x) to it, do that a whole bunch of times and take the mean. Do that for each term and you've got an estimate.

    Notice that if all the r.v.s have a common mean then each of the F_i(x) is the same and each of the f_i(x) are the same so this simplifies to
    P(X_1 >max\{X_2,\ldots,X_n\}) = \1 - (n-1) \int_{-\infty}^\inftyf_o(x)(F_o(x))^{n-1}dx where f_o(x) is the common pdf

    but \int_{-\infty}^\inftyf_o(x)(F_o(x))^{n-1}dx = \frac{1}{n} so the whole expression is equal to 1/n which matches our intuition.

  2. #17
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    Hello

    I started a new thread here:

    http://talkstats.com/showthread.php?t=12661

    I think we 're talking about the same problem.
    My link appears to have the solution but I cannot test it because I don't use "Mathematika".

    But with pairwise comparisons I think one gets nowhere.

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    AHA! I think I managed the deriviation of your result with the help of the article above and of course your end result..
    A moment please, and I shall post it here and you can let me know if I did it right

    ...

    Given a set of nonidentically distributed random varialbes X_0, X_1, X_2,\ldots,X_n what is the probability that X_i is the maximum value.
    P[X_i>Y_i] where Y_i = max(X_{i'}|_{\forall i'\neq i})
    P[X_i>Y_i]=P[X_i-Y_i>0]=1-F_{X_i-Y_i}(0) where F_{X_i-Y_i}(0) is the CDF of X_i-Y_i

    Take the convolution.. F_{X_i-Y_i}(0)=\int_{-\infty}^{\infty} F_i(0-(-y))\cdot f_y(y)dy=\int_{-\infty}^{\infty} F_i(y)\cdot f_y(y)dy where F_i is the CDF of X_i
    f_y(y) is the PDF of Y so we can calculate Y's CDF and differentiate.

    P[Y<a] = P[X_{i'}<a]\cdot P[X_{i'+1}<a]\cdots = \prod_{\forall i'\neq i}P[X_i'<a] = F_y(a)
    F_y(y)=\prod_{\forall i'\neq i}F_i(y)
    f_y(y)=\frac{d}{dy} F_y(y)= \frac{d}{dy}F_i \cdot F_{i+1} \cdots = f_i \cdot F_{i+1} \cdot F_{i+2} \cdots + F_i \cdot f_{i+1} \cdot F_{i+2} \cdots + \cdots

    ok so my notation is a bit crazy... but I think it turns out the same as yours, Dason.
    I'm just wondering now if there is a sneakier way to do it. I suppose it would be different depending on the distributions used for the random variables. I could take a hint from the article posted for the normal distributions..
    Last edited by mightymergz; 07-09-2010 at 05:06 PM.

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    Looks pretty good. At first I thought of doing the convolution to look at the difference between X_i and Y_i but then decided to just look at the double integral. Turns out pretty nice either way it looks like.

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    Wonderful thanks guys!

    But now I want to complicate the problem some more

    Consider the same problem but now the variables aren't independent but are correlated. Say that the correlation is described by some function A(\Delta t) where the correlation between two variables only depends on the distance between them. i.e. the \Delta t for variables X_i, X_j would be j-i

    I'm going to start thinking about it now. If anyone has any insights they would be greatly appreciated!

  6. #21
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    I think I came up with a simplified form of the solution..

    We can rearrange the convolution because it is commutative. Thus we can avoid having to calculate the nasty pdf of Y f_Y(x)
    F_{X-Y}(a)=P[X-Y\le a]

    =\iint\limits_{x-y\le a} f_X(x)f_Y(y)dxdy

    =\int_{-\infty}^{\infty}f_X(x)dx \int_{x-a}^{\infty} f_Y(y)dy

    =\int_{-\infty}^{\infty}\left\{f_X(x)dx \cdot [1-F_Y(x-a)]\right\}

    =\int_{-\infty}^{\infty}f_X(x)dx - \int_{-\infty}^{\infty}F_Y(x-a)f_X(x)dx

    =1 - \int_{-\infty}^{\infty}F_Y(x-a)f_X(x)dx

    Thus P[X-Y>0]=1-\left(1-\int_{-\infty}^{\infty}F_Y(x)f_X(x)dx\right)

    P[X-Y>0]=\int_{-\infty}^{\infty}F_Y(x)f_X(x)dx

    where F_{Y_i}(x)=\prod_{i'\neq i}F_i(x) i.e. the product of the cdf's for all the rv's except the one for which we want the probability of it being the maximum.

    and f_{X_i}=f_i(x) which is the pdf of the rv we are interested in.

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