$P(X_{1}+1>X_{2})$ for exponential variables

Question

Let $X_{1}, X_{2}$ be independent random variables having exponential distribution with $\lambda = 1$. Find the density of $X_{1} + X_{2}$ and calculate $P(X_{1}+1>X_{2})$.

If I'm not mistaken the answer to the first question should be $te^{-t}$ where $t$ is non-negative (using convolution). However, I'm struggling with the other task, how could I possibly use the result I've already obtained to solve it?

Answer 1

Given that $X_1$ and $X_2$ are independent, with probability distribution functions $$f_{X_1}(x_1) = e^{-x_1} \\ f_{X_2}(x_2) = e^{-x_2}$$ with $x_1,x_2 > 0. $
Then we know that the joint probability density function of $(X_1,X_2)$ is $$f_{X_1,X_2}(x_1,x_2) = f_{X_1}(x_1)f_{X_2}(x_2) \qquad \text{by independence} \\ = e^{-x_1}e^{-x_2}.$$
To evaluate $\mathbb{P}(X_1 + 1 > X_2$), this is equal to the integral of the joint density function, with integral bounds reflecting the probability. That is, $x_1 > x_2 -1$,
$$\mathbb{P}(X_1 + 1 > X_2) = \int_{0}^{\infty}\int_{0}^{x_1 + 1} e^{-x_1}e^{-x_2}dx_2 dx_1$$

To see the bounds clearly, try drawing a region in the $x_1,x_2$ plane, satisfying $x_1 > x_2 + 1$ and $x_1,x_2>0$.

Answer 2

Alternative solution:

$$P(X_1+1>X2)=$$$$P(X_1+1>X_2\mid X_2\leq 1)P(X_2\leq1)+P(X_1+1>X_2\mid X_2> 1)P(X_2>1)=$$$$P(X_2\leq1)+\frac12P(X_2>1)=1-e^{-1}+\frac12e^{-1}=1-\frac12e^{-1}$$

Here the second equality is based on the fact that exponential distribution is "memoryless".

Together with $X_1,X_2$ being iid with continous distribution that gives:$$P(X_1+1>X_2\mid X_2> 1)=P(X_1>X_2-1\mid X_2>1)=P(X_1>X_2)=\frac12$$

The distribution of $X_2-1$ under condition $X_2>1$ is the same as the distribution of $X_2$.