Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Function of a random variable and conditioning

  1. Oct 11, 2011 #1
    Ok, since nobody answered my last problem, I simplify. :)

    Let Z = γ1X1 + γ2X2, where the gammas are just constants
    p(Z) = exp(Z)/(1 + exp(Z))
    X1 and X2 are bivariate normal and put
    Y = α + β1X1 + β2X2 + ε where ε ~ N(0,σ).

    Now, we want to find f(p(Z)|X1,Y). In this case, is it legal to do the
    operation f(p(Z)|X1,Y)=f(p(Z|X1,Y))?

    That is can we write
    f(exp(Z|X1,Y)/(1 + exp(Z|X1,Y)))?

    Thanks for any help!
    /H
     
  2. jcsd
  3. Oct 11, 2011 #2

    Stephen Tashi

    User Avatar
    Science Advisor

    Let [itex] X_1 [/itex] and [itex] X_2 [/itex] each be bivariate normal random variables
    Let [itex] Z = \gamma_1 X_1 + \gamma_2 X_2 [/itex] where [itex] \gamma_1 [/itex] and [itex] \gamma_2 [/itex] are constants.
    Let [itex] Y = \alpha + \beta_1 X_1 + \beta_2 X-2 + \epsilon [/itex] where [itex] \alpha,\beta_1,\beta_2 [/itex] are each constants and [itex] \epsilon [/itex] is a normal random variable with mean 0 and standard deviation [itex] \sigma [/itex].

    Is that notation supposed mean you want the probability density function for [itex] Z [/itex] given [itex] X_1 [/itex] and [itex] Y [/itex] ?

    I don't know what that notation means. The conditional density function of [itex] Z [/itex] is some function of the variables [itex] Z, X_1,Y [/itex] but what does the notation "exp(Z|X1,Y)" mean?
     
    Last edited: Oct 11, 2011
  4. Oct 11, 2011 #3
    Yes.


    For instance, if we want to know the distribution of p(Z) = exp(Z)/(1 + exp(Z)) and we know the distribution of Z, then we make a simple transformation, put the inverse in the pdf of Z and multiply with the derivative as usual.

    However, the problem is finding the disitrbution of p(Z)|Y. My idea was then to put the inverse in the pdf of Z|Y and then multiply with the inverse. I am not sure if my approach is correct, but if you have another suggestion of how to proceed I would be grateful.

    /H
     
  5. Oct 11, 2011 #4

    Stephen Tashi

    User Avatar
    Science Advisor

    Let's try again:

    Let [itex] X_1 [/itex] and [itex] X_2 [/itex] be random variables that have a joint bivariate normal distribution (rather than each of them being bivariate normal).
    Let [itex] Z = \gamma_1 X_1 + \gamma_2 X_2 [/itex] where [itex] \gamma_1 [/itex] and [itex] \gamma_2 [/itex] are constants.
    Let [itex] W = \exp(Z)/(1 + \exp(Z)) [/itex]
    Let [itex] Y = \alpha + \beta_1 X_1 + \beta_2 X_2 + \epsilon [/itex] where [itex] \alpha,\beta_1,\beta_2 [/itex] are each constants and [itex] \epsilon [/itex] is a normal random variable with mean 0 and standard deviation [itex] \sigma [/itex].

    Do you want the conditional distribution of [itex] W [/itex] given [itex] X_1 [/itex] and [itex] Y [/itex] ? (Your other thread mentioned a joint distribution instead of conditional distribution and also it mentioned that the final goal was to find an expected value.)
     
  6. Oct 11, 2011 #5
    The final goal is to find the conditional distribution of X1 and Y given W. Of course, there are different ways of getting there depending on how you calculate the joint X1, Y, W.

    My question in this thread may solve a part of the problem and also the disitrbution of W given X1 and Y.
     
  7. Oct 12, 2011 #6

    Stephen Tashi

    User Avatar
    Science Advisor

    Now that the problem is established, help me understand the question about technique.

    Who's inverse and who's derivative are you talking about? Let's say [itex] Z [/itex] has cumulative distribution [itex] F_Z(x) [/itex] with inverse function [itex] {F_Z}^{-1}(x) [/itex]. Using that notation, what is your claim about the probability density (or cumulative distribution) of [itex] W [/itex] ?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook