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!!!

  • $\begingroup$ For this kind of question you should add the self-study tag and read it's wiki. $\endgroup$ – Andy Jul 29 '15 at 8:50
  • $\begingroup$ By considering the case $n=1$ it should be immediately obvious that the conclusion is false, for then $(1/n)\sum X_iI_i$ has a finite upper bound of $\mu$. This observation translates to all $n$: the weighted sum has an upper bound of $n\mu$ and therefore cannot possibly be Normal. $\endgroup$ – whuber Jul 29 '15 at 12:17

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.

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  • $\begingroup$ As far as I can see, $I$ is a random variable, and the question is whether or not the variable $Z=X\cdot I$ is a normal random variable, which it is not. (EDIT: this is a response to a now deleted comment...) $\endgroup$ – yoki Jul 29 '15 at 9:00
  • $\begingroup$ $I$ is just an indicator... The general statement is actually: $X_{1},\ldots X_{n} \sim N(\mu, \sigma^{2})$ and 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} And I have to derive that $\frac{1}{n}\sum\limits_{i=1}^{n}X_{i}I_{i}$ is normally distributed. (The statement is in one book and I have to write down the proof) That's why $XI$ should be normally distributed as well.....I wanted to proof it for the entire sum with the induction... $\endgroup$ – Alia Jul 29 '15 at 9:16
  • $\begingroup$ The indicator is a random variable, as yoki said. It's $\text{Bernoulli}(\frac12)$ distributed. Further, $X_iI_i$ is clearly not normal, but a flipped, shifted half-nomal + a spike at $0$. $\sum X_iI_i$ is therefore not normal. $\endgroup$ – Glen_b -Reinstate Monica Jul 29 '15 at 9:40
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    $\begingroup$ @Alia , the probability for each $X_i$ to be higher than its mean is $0.5$. The overall sum will be zero if each of the $X_i$ is higher than its mean, and the probability for this event is neither zero nor one (unless they exhibit specific types of dependencies), so it cannot be a Gaussian variable. $\endgroup$ – yoki Jul 29 '15 at 9:48
  • $\begingroup$ Guys, thanks a lot for your comments. I really approciate your help... I understood why the way I was thinking is not correct... But...Is there any way to show that $1/n*\sum X_{i]I_{i}$ is at least asymptotically normally distributed? ....maybe whith the central limit theorem? $\endgroup$ – Alia Jul 29 '15 at 14:05

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