Normal Distribution

Question

I am a little bit confused about the correctness of the following statement. Could you check please, if my small "proof" is correct?

Let $X_{1}, \ldots, X_{n}$ be a i.i.d sequence that follows a normal distribution $X_{i}\sim N(\mu, \sigma^{2})$ and $I_{i}$ an indicator with:

\begin{equation*} I_{i}= \begin{cases} 1 & \text{if} X_{i}<\mu\\ \frac{1}{2} &\text{if} X_{i}=\mu\\ 0 & \text{if} X_{i} > \mu \end{cases} \end{equation*}.

I have to show that $\frac{1}{n}\sum\limits_{i=1}^{n}X_{i}I_{i}$ is normally distributed.

I had the idea to prove it with the induction:

for n=1: Since

\begin{align*} XI=\begin{cases} X & \text{if } X<\mu\\ \frac{1}{2}\mu &\text{if} X=\mu\\ 0 & \text{if } X > \mu \end{cases} \end{align*}

and $\mu, X\sim N(\mu, \sigma^{2})$ $XI$ is normally distributed.

Then the similar derivation for n=n+1

Is this way correct?

Thank you so much in advance!!!

Answer 1

First of all, since $X$ is a continuous random variable, the probability $P\{X=\mu\}=0$.

Second, this statement is not correct, because a necessary condition for a normal random variable is that it is continuous, and it should satisfy $P\{X=x\}=0$ for all $x$. In your case, $P\{XI=0\}=P\{X>\mu\}=0.5$, so it is definitely not a Gaussian.

Note that a Gaussian can be degenerate, where $\sigma=0$, and then $P\{X=\mu\}=1$, but it is not true in general.