Compute $P\{t\lt X\lt Y\}$ for $t \ge 0$, when $X$ and $Y$ are independent exponential random variables, possibly with different parameters

Question

What is $P(t \lt X \lt Y)$ for $t\gt 0$, when X and Y are independent exponential variables with the parameters $\lambda$ and $\mu$ respectively?

Since X and Y are independent, I got joint PDF to be $\lambda \cdot\mu\cdot e^{-\lambda\cdot x-\mu\cdot y}$.

Now that I'm looking for $P(t \lt X \lt Y)$, using the joint PDF, I set it as: $$ P(t<X<Y)=\int_0^\infty\int_t^y \lambda \cdot\mu\cdot e^{-\lambda\cdot x-\mu\cdot y}dxdy $$

Is this correct? I'm trying to solve it using the online integral calculator but it's not getting the answer.

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Edit 1: Here are the changes and result I came up with

Edit 2: Fixed my calculation error and got a confident answer: $$ P(t<X<Y)=\int_t^\infty\int_t^y \lambda \cdot\mu\cdot e^{-\lambda\cdot x-\mu\cdot y}dxdy=\frac{\lambda}{\lambda+\mu}e^{-t(\lambda+\mu)} $$

I would really appreciate if someone could check if this is correct. Thank you!

Answer 1

Try this in Mathematica

Probability[ t < x < y, {x \[Distributed] ExponentialDistribution[\[Lambda]], y \[Distributed] ExponentialDistribution[\[Mu]]}]

You will get $$ \begin{array}{cc} & \left\{ \begin{array}{cc} \frac{\lambda }{\lambda +\mu } & t<0 \\ \frac{\lambda e^{-t (\lambda +\mu )}}{\lambda +\mu } & t \ge 0 \\ \end{array} \right. \\ \end{array} $$