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I have two variables $X_1$ and $X_2$ are IIDs distributed with the common distribution $X \sim Exp(\lambda)$ for some $\lambda >0$. I was working to find a distribution of $|X_1−X_2|$.

I used convolution to find the function $f_Z(z)$ as \begin{align} f_Z(z) &= \int_{-\infty}^{\infty}f_x(x)*f_y(x-z)dx , \ for \ X_1 > X_2 \ \& \\ f_Z(z) &= \int_{-\infty}^{\infty}f_{-x}(x)*f_y(z+x)dx , \ for \ X_1 < X_2 \end{align}

Is this the correct method to do it & how to find the range of $z$?

Alessio K
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Abijith P Y
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1 Answers1

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By symmetry between $X_1$ and $X_2$ we can write $$P(|X_1-X_2| \leq z)$$ $$=2P(X_2 \leq X_1\leq X_2+z)$$ $$=2\int_0^{\infty}\int_{x_2}^{x_2+z} \lambda^{2}e^{-\lambda x_1}e^{-\lambda x_2}dx_1dx_2.$$ for $z>0$. I will let you do the computation.

Kavi Rama Murthy
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  • I didn't understand how the lower limit of x1 became x2 – Abijith P Y Sep 04 '20 at 06:36
  • @AbijithPY You have to integrate the joint density over the region ${(x_1,x_2): x_2 \leq x_1 \leq x_2+z}$. – Kavi Rama Murthy Sep 04 '20 at 06:38
  • Because $$\begin{align}P(\lvert X_1-X_2\rvert\leqslant z)&=2,P(X_2\leqslant X_1\cap X_1-X_2\leqslant z)\&=2,P(X_2\leqslant X_1\leqslant X_2+z)\&=2\int_0^\infty f_{X_2}(x_2),P(x_2\leqslant X_1\leqslant x_2+z),\mathrm d x_2\&=2\int_0^\infty\int_{x_2}^{x_2+z} f_{X_2}(x_2),f_{X_1}(x_1),\mathrm d x_1,\mathrm d x_2\end{align}$$ – Graham Kemp Sep 04 '20 at 07:58
  • Okay thanks got it – Abijith P Y Sep 04 '20 at 11:58