Expectation with Dirac measure

Question

Let $\mathbf{P}_X [A] := \frac{1}{2} \delta_0 (A) + \frac{1}{2} \int_{A \cap (0, \infty)} e^{-t} dt$ for $A \in \mathcal{B} (\mathbb{R})$. What is $\mathbf{E} [X]$?

I tried finding the density and computed $\mathbf{E} [X]$ as follows:

$F_X (x) = \mathbf{P} [X \le x] = \mathbf{P}_X [(-\infty, x]] = \tfrac{1}{2} \delta_0 ((-\infty, x]) + \frac{1}{2} (1 - e^{-x}) \mathbf{1}_{(0, \infty)} (x)$;

$f_X (x) = \begin{cases}\frac{1}{2} e^{-x}, & x > 0; \\0, & \text{else}.\end{cases}$

Hence, $\mathbf{E} [X] = \int_0^\infty x \cdot \frac{1}{2} e^{-x} d x = \frac{1}{2}$.

Is this solution correct? Is there a better approach?

Answer 1

Without resorting to densities (which as pointed out aren't entirely rigorous in this case since $P(X=0)=1/2$) you can use the fact that your random variable is non-negative and hence $$ EX=\int_0^\infty 1-F(x)\, dx=\int_0^\infty \frac{e^{-x}}{2}\, dx=\frac{1}{2}. $$

Answer 2

Your approach gives the right answer but isn't rigorous because the distribution doesn't have a true probability density function.

A way to make your argument rigorous is to prove the following result: Suppose $\mu$ and $\nu$ are measures and that $f$ is both $\mu$ and $\nu$ integrable. Prove that $\pi(A) := \mu(A) + \nu(A)$ is a measure, $f$ is $\pi$-integrable, and $$ \int f(x) \ \pi(\text{d}x) = \int f(x)\ \mu(\text{d}x) + \int f(x)\ \nu(\text{d}x) .$$ This is a rather intuitive result, the easiest way I can think of to prove it is to use the monotone class theorem for functions. Then just exploit that $P_X(A) = \frac{1}{2} \delta_0(A) + \frac{1}{2} \mu(A)$ where $\mu$ is the measure corresponding to the exponential distribution.