# Conditional density for bivariate logistic law

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

## 1 Answer

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.

• 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 – Brocolus Mar 4 '18 at 22:43