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Say X and Y are two independent random variables with exponential density\begin{split} f_{X}(x) = a e^{-ax}\end{split} and \begin{split} f_{Y}(y) = b e^{-by}\end{split}, then what is the probability density function of Z=X-Y? I'm trying to slove this problem, but I have no sense of the integrating regions. How will they be?

I tried Shai's hint \begin{split} {\rm P}(X - Y \le z) = \int_0^\infty {{\rm P}(X - Y \le z|Y = \tau )f_{Y}(\tau ){\rm d}\tau } = \int_0^\infty {{\rm F_{X}}(z + \tau )f_{Y}(\tau ){\rm d}\tau } \end{split}

I obtained this \begin{split} {\rm P}(Z \le z) = F_{Z}(z) = 1 - \frac b {a+b} e^{-a z} \end{split} But it's not converage to 1 when z is infinite, what's wrong with my calculation?

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    You should separate into the cases $z + \tau < 0$ and $z + \tau > 0$.2011-04-11
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    Also, it does converge to $1$ when $z \to \infty$; the problem is seen when $z \to -\infty$.2011-04-11
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    One may a priori expect to find $|z|$ in the answer.2011-04-11
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    Thank you! sorry for the mistake ,but how to obtain |z| in the answer?2011-04-11
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    Have you calculated the distribution function for all $z \in \mathbb{R}$? Note that $Z$ takes values in $(-\infty,\infty)$. You can use $|z|$ to shorten the final answer.2011-04-11
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    I don't understand, X=Z+Y, when z < -y \begin{split} F_{X}(z+y) \end{split}is out of the domain of cdf \begin{split} F_{X} \end{split}how to split the integration regions?2011-04-11
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    Suggestion: compute $F_Z (z) = \int_0^\infty {F_X (z + \tau )f_Y (\tau )d\tau } $ for $z > 0$ and $z < 0$ separately. The case $z > 0$ should be very easy, since then $z + \tau > 0$ as well. For the case $z < 0$, recall my first comment above.2011-04-11
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    in OK's case, how would the integral limits go? For z>0 and z<0 should you get the same distribution? I'm looking at 1 - (1/2)ae^-az for the z>0 case.2011-04-12

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Hint: $P(X - Y \le z) = \int_0^\infty {P(X \le z + y)f_Y (y)dy}$, $z \in \mathbb{R}$.

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    So, compute the right-hand side and differentiate.2011-04-11
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    Where exponential random variables are concerned, one can prefer considering $P(X\ge t)$ rather than $P(X\le t)$, in which case the quantity to introduce would be $P(X-Y\ge z)$ rather than $P(X-Y\le z)$.2011-04-11
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    Of course, but it might be more convenient (to the OP) to work with the distribution function rather than the tail distribution function. The difference is very little anyway.2011-04-11
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I believe that it's eventually the same. But you can also do this $$P(X-Y\le z)=\iint_{\{x-y\le z\}}f_{X,Y}(x,y)dxdy$$ where the joint PDF is given by $f_{X,Y}(x,y)=f_X(x)f_Y(y)$ because of independence.