# Define random variable distribution given standard normal random variable

I have a homework but I don't know how to solve it or what should I do. Kindly help me or guide me.

Let a constant $a$ satisfy that $\int_{0}^{a}{x^2 e^{-\frac{x^2}{2} }dx} = \int_{a}^{\infty}{x^2 e^{-\frac{x^2}{2} }dx}$

Suppose $X$ is a standard normal random variable. Define $Y$ as follows

$$Y = \left\{ \begin{array}{ll} X & \quad if|X|\geq a \\ -X & \quad if|X| < a \end{array} \right.$$

(a) What is the distribution of $Y$?

(b) Show that $X$ and $Y$ are uncorrelated.

(c) Show that $X$ and $Y$ are not independent.

Actually I don't understand the purpose of given integration function, like how or what can I do with the function. What I understand so far (maybe true or false), I consider when $|X| \geq a$ that's mean $Y = \int_{a}^{\infty}{x^2 e^{-\frac{x^2}{2} }dx}$ and the opposite when $|X| < a$. If I consider this, that's mean Y is uniform distribution because nothing different when |X| is greater or less than $a$. Then to show X and Y are uncorrelated, $Cov(X,Y)=E[XY]-E[X]E[Y]=0$ that means $E[XY]=E[X]E[Y]$. But this understanding is the opposite of (c) question, since this apply when X and Y are independent. Thank you so much for your help.

• Please add the [self-study] tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. – gung - Reinstate Monica Sep 19 '16 at 16:41
• Thank you so much for your guidance. What I understand is, because the integration value is given I consider when $|X| \geq a$ that's mean $Y = \int_{a}^{\infty}{x^2 e^{-\frac{x^2}{2} }dx}$ and the opposite when $|X| < a$. If I consider this, that's mean Y is uniform distribution because nothing different when |X| is greater or less than a – Jyanto Sep 20 '16 at 12:56
• Independence and correlation are different things. Independence has to do with mutual information, and correlation has to do with a specific linear relationship. Draw a scatter-plot of $X$ and $Y$, and that may give you some intuition as to how the correlation might be balanced at 0, while there still is a lot of mutual information. – Matthew Graves Sep 20 '16 at 15:39
• (To be clearer, "mutual information" is a measure of how well I could predict $X$ given $Y$, and how well I could predict $Y$ given $X$. Independence is the statement that there's no mutual information, that is, my ability to predict $Y$ or $X$ is the same as it was before you told me the other one.) – Matthew Graves Sep 20 '16 at 15:40

## 1 Answer

(a) What happens when you multiply a symmetric distribution by -1? That is, what's the difference between the distribution of $X$ and $-X$?

(b) If two random variables are uncorrelated, then that implies that their covariance is 0. Covariance is $E[(X-E[X])(Y-E[Y])]$, which is made simpler by the fact that $E[X]=E[Y]=0$. What happens when you integrate that across the domain of the function? (The definition of $a$ should come in handy.)

(c) Independence implies that $P(X=x, Y=y)$ factorizes to $P(X=x)P(Y=y)$. Can you exhibit a pair $(x, y)$ where that doesn't hold?

• Thanks Matthew. I still don't get the idea. (a) how could I know that was symmetric distribution? Is it because X is normal distribution (it looks like a bell) or because of the given integration function. if X is bell on the positive x-axis, -X is bell on the negative x-axis? – Jyanto Sep 20 '16 at 13:06
• Jyanto, it's because whenever $x$ is used, it's always $x^2$, which is symmetric around 0. (All polynomials with only even powers are.) – Matthew Graves Sep 20 '16 at 15:34
• (But yeah, you also could have gotten that from just knowing that the normal distribution was symmetric, and not why it's symmetric.) – Matthew Graves Sep 20 '16 at 15:44
• So you know the X is symmetric from the integration function? since the X is symmetric when positive and negative, that the Y will become symmetric as well? So can I conclude Y is normal distribution as well because Y is function of X? – Jyanto Sep 20 '16 at 15:50
• Jyanto, sorry, that was a little misleading. $X$'s distribution is a standard normal, which is proportional to $e^{-x^2}$, which is what makes $X$ symmetric. $Y$, viewed by itself, is also a standard normal distribution, but it's worth stepping through how the partial reflection of a symmetric distribution works. – Matthew Graves Sep 20 '16 at 17:44