An Example on Mixture Distribution

Question

Let $U$ be a uniform random variable on $(0, 1)$ and $Y$ be a Bernoulli random variable with $0.5$ as parameter. In addition, $U$ and $Y$ are independent. Now, define $X:= U+Y$. I think $X$ is a uniform random variable on $(0, 2)$. Is this conclusion correct, please? If so, how do I justify it in a formal way, please? Thank you!

Answer 1

Yes, you are correct. If two distributions have the same moment-generating function (MGF), then they are identical at almost all points and have the same distribution function $F(x)$ for all $x$. Also, the moment-generating function of a sum of random variables is the product of the moment generating function of the variables.

The MGF for the Bernoulli with parameter $p, (q = 1 - p)$ and the continuous uniform with parameters $a, b$are: $$ \begin{align} MGF_B &= pe^t + q\\ MGF_{CU} &= \frac{e^{tb} - e^{ta}}{t(b - a)} \end{align} $$

In our specific case we have $p = 0.5, a = 0, b = 1$, so: $$ \begin{align} MGF_B &= \frac{1}{2}e^t + \frac{1}{2} = \frac{e^t + 1}{2}\\ MGF_{CU} &= \frac{e^{t} - 1}{t} \end{align} $$

The product of the two is: $$ \frac{e^{2t} - 1}{2t} $$

This is just the MGF of the continuous uniform with $a = 0$ and $b = 2$, so the the two have the same distribution.