What is the Joint Distribution for a Bivariate Normal and Logistic Distribution?

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The discussion focuses on deriving the joint distribution for a bivariate normal and logistic distribution. The user seeks help with the univariate case where X follows a normal distribution and Y is defined as a logistic transformation of X. They express uncertainty about the joint probability function f(x,y) and the expected value E(XY), questioning whether the probability mass can be concentrated at a single point. The bivariate case extends this concept with two normal variables, X1 and X2, and a logistic transformation involving constants. The user is looking for guidance on how to proceed with these calculations.
Hejdun
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Hi!

I would be really happy to receive some help. I have tried using Jacobians and so on, but
I am stuck.

I'll start with the univariate case. Let X ~ N(μ,σ) and Y = exp(X)/(1+exp(X)). What is the joint f(x,y)? According to intuition fy|x = 1, but since we are dealing with continuous distributions I am not sure.

The bivariate case is an extension. Let X1, X2 be bivariate normal and
Y = exp(γ1*X1 + γ2*X2)/(1 + exp(γ1*X1 + γ2*X2) ), where the gammas are just
constants. Now I am looking for f(x1,x2,y). Any suggestions of how to proceed at least?

/H
 
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Hejdun said:
Hi!

I would be really happy to receive some help. I have tried using Jacobians and so on, but
I am stuck.

I'll start with the univariate case. Let X ~ N(μ,σ) and Y = exp(X)/(1+exp(X)). What is the joint f(x,y)? According to intuition fy|x = 1, but since we are dealing with continuous distributions I am not sure.

The bivariate case is an extension. Let X1, X2 be bivariate normal and
Y = exp(γ1*X1 + γ2*X2)/(1 + exp(γ1*X1 + γ2*X2) ), where the gammas are just
constants. Now I am looking for f(x1,x2,y). Any suggestions of how to proceed at least?

/H


Just to clarify the first case. I actually want to find the joint f(x,y) and then E(XY). Since Y=g(X), then f(x, g(x)) and E(XY)= E(Xg(X)). We know that Pr(Y=exp(x)/(1 + exp(x))) = 1, but I am not sure if then can write fy|x = 1 (all probability mass concentrated in one point). If that is the case, then f(x,y) = f(x) and E(XY) = ∫∫xyf(x)dxdy = ∫xexp(x)/(1+exp(x))f(x)dx which can easily be calculated numerically. Am I on the right track?

The bivariate case is more complicated but maybe that can be solved as well. :)

/H
 
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