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.

  • 2
    $\begingroup$ 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. $\endgroup$ – gung - Reinstate Monica Sep 19 '16 at 16:41
  • $\begingroup$ 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 $\endgroup$ – Jyanto Sep 20 '16 at 12:56
  • $\begingroup$ 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. $\endgroup$ – Matthew Graves Sep 20 '16 at 15:39
  • $\begingroup$ (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.) $\endgroup$ – Matthew Graves Sep 20 '16 at 15:40

(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?

  • $\begingroup$ 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? $\endgroup$ – Jyanto Sep 20 '16 at 13:06
  • $\begingroup$ 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.) $\endgroup$ – Matthew Graves Sep 20 '16 at 15:34
  • $\begingroup$ (But yeah, you also could have gotten that from just knowing that the normal distribution was symmetric, and not why it's symmetric.) $\endgroup$ – Matthew Graves Sep 20 '16 at 15:44
  • $\begingroup$ 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? $\endgroup$ – Jyanto Sep 20 '16 at 15:50
  • $\begingroup$ 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. $\endgroup$ – Matthew Graves Sep 20 '16 at 17:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.