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I'm working on an exercice of my probability book but which doesn't provide any solution or hint, I'd like to share my research if you could give me a hand to finish it :

let a vector $X=(X_1,X_2)$ and let an introduced bivariate version of logistic law introduced by Gumbel for joint distribution function : $F_{\textbf{X}}(x_1,x_2)=\dfrac{1}{1+\exp(-x_{1})+\exp(-x_{2})}$ for $x \in \mathbb{R^{2}} $

What is aked : Demonstrate that conditional density of $X_1$ knowing $X_2=x_2$ is : $f_{1|2}(x_1|x_2)=\dfrac{2\exp(-x_1)(1-\exp(-x_2))^2}{(1+\exp(-x_1)+\exp(-x_2))^3}$

What I did : As $X$ is continuous variable, I'll have to calculate the margin density of the problem. Therefore I'll work on : $f_{1|2}(x_1|x_2)=\dfrac{f_\textbf{X}(x_1,x_2)}{f_2(x_2)}$

Then, I go for calculating each member i find : $f_\textbf{X}(x_1,x_2)=\dfrac{\exp(-x_1)+\exp(-x_2)}{(1+\exp(-x_1)+\exp(-x_2)^2}$ $f_2(x_2)=\dfrac{\exp(-x_2)}{(1+\exp(-x_2))^2}$

Then, i find : $f_{1|2}(x_1|x_2)=\dfrac{\dfrac{\exp(-x_1)+\exp(-x_2)}{(1+\exp(-x_1)+\exp(-x_2)^2}}{\dfrac{\exp(-x_2)}{(1+\exp(-x_2))^2}}\\=\dfrac{\exp(-x_1)+\exp(-x_2)}{\exp(-x_2)}\cdot\dfrac{(1+\exp(-x_2))^2}{(1+\exp(-x_1)+\exp(-x_2))^2}=\dfrac{(1+\exp(-x_1)+\exp(-x_2))(\exp(-x_1)+\exp(-x_2))(1+\exp(-x_2))^2}{\exp(-x_2)(1+\exp(-x_1)+\exp(-x_2))^3}=\dfrac{(\exp(-x_1)+\exp(-x_2)+(\exp(-x_1)+\exp(-x_2))^2)(1+\exp(-x_2))^2}{\exp(-x_2)(1+\exp(-x_1)+\exp(-x_2))^3}$

From this point, I'm stuck to find the result I have to find, Is my method right? I'm struggling with this one for a little while, but I'm wondering how to get the "$-$" sign when every term is positive

Thank you in advance for your reading and your help

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You computed the joint density incorrectly. $$f(x_1, x_2) = \frac{\partial^2}{\partial x_1 \partial x_2}F(x_1, x_2) = \frac{e^{-x_1 - x_2}}{(1+e^{-x_1}+e^{-x_2})^3}$$ $$f_2(x_2) = e^{-x_2} \int e^{-x_1}(1+e^{-x_1} + e^{-x_2})^{-3} \, dx_1 = -e^{-x_2} \int_0^\infty (1 + u + e^{-x_2})^{-3} \, du = \frac{1}{2} e^{-x_2} (1+e^{-x_2})^{-2}$$ Then $f(x_1,x_2)/f_2(x_2)$ yields the desired answer.

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  • $\begingroup$ Thank you very much for your answer ! The exercice was offering to use the formula $f_X=F_X(x)(1-F_X(x))$ to solve the exercice but it led me nowhere. Thank you again ! But the "-" in the numerator must be a mistake then $\endgroup$ – Brocolus Mar 4 '18 at 22:43

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