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I recently had to solve this same problem, except $X$ and $Y$ were uniform on $[0,1]$. The joint probability distribution was uniform, so I just needed to find the proportion of the area inside the region of the unit square where $|X-Y|\leq z$ for a given $z$.

I was wondering: what should I do if $X$ and $Y$ are independent exponential random variables with parameter $\lambda$? I believe I would have to integrate $$\int_{y_0}^{y_1} \int_{x_0}^{x_1} \lambda^2 e^{-\lambda(x+y)} \, \mathrm dx \, \mathrm dy$$

but I'm not sure what the bounds would be. How should one approach this?

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You can use: $$ P(\lvert X - Y \rvert < z ) = P(X - Y < z, X - Y > -z) = P(X - Y < z) - P(X - Y < z, X - Y < -z) = P(X- Y < z) - P(X-Y<-z)$$

While $X-Y$ has pdf that's quite easy to calculate - see another answer. Moreover, it's form allows us for closed-form expression for cdf. Said PDF is:

$$ f(x) = \frac{\lambda \mu}{\lambda+\mu} \cases{e^{-\mu x} & if $x > 0$\cr e^{\lambda x} & if $x < 0$\cr}$$

where $\mu, \lambda$ are parameters of $X, Y$ respectively.

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