How to prove dependence of random variables

by Mari   Last Updated May 16, 2019 03:19 AM

I need to solve the following problem.

Let $X$ be a normal random variable with mean $\mu$ and standard deviation $\sigma$ and let $I$, independent of $X$, be such that $\mathbb{P}(I = 2) = \mathbb{P}(I = -2) = 0.5$. Let $Y = I X$. In words, $Y$ is equally likely to be either $2X$ or $–2X$.

a) Are $X$ and $Y$ independent?

b) Are $I$ and $Y$ independent?

c) What is the distribution of $Y$?

d) Find $\mathrm{Cov}(X,Y)$.

I think $X$ and $Y$ are dependent, but I don't know how to prove it. Any hint? Also, any hint about the following points?

Answers 3

Hint about Question 1.

The intuitive notion behind independence of $X$ and $Y$ is that knowledge of the value of one random variable $X$ should provide you with no information about the value of $Y$. So start with basics. Is it true that $Y$ can take on all real number values? If you are told that $X$ has value $1$, does it continue to be true that $Y$ can take on all real number values? If your answer is YES, re-read your own assertion

In words, $Y$ is equally likely to be either $2X$ or $–2X$

at least three times, and think some more about what this is saying when you know that $X=1$.

Hint for Question 2.

Repeat the above argument with $I$ in place of $X$. What is the conditional mean of $Y$ when $I=2$? What about when $I = -2$?

Dilip Sarwate
Dilip Sarwate
October 19, 2014 22:46 PM

Here is my thoughts. If $X$ and $Y$ are independent, then we have $\mathbb{P}(Y = 2 \mid X = 1) = \mathbb{P}(Y = 2)$. So we can check whether they are equal.

On the one hand, $$P(Y = 2 \mid X = 1) = P(IX = 2 \mid X = 1)$$

$$ = P(I = 2, X = 1 \mid X = 1) = P(I = 2) = 0.5$$

On the other hand, $$P(Y = 2) = 0.5*P(X=1) + 0.5*P(X=-1) = 0.$$ (NOT sure whether I am correct here.) Thus I think they are dependent.

Whether $I$ and $Y$ are independent can also be checked by the same fashion.

As for the distribution of $Y$, $$F_Y(y) = P(Y <= y)=P(2X <= y \mid I=2)*P(I=2)$$

$$ + P(-2X <= y \mid I=-2)*P(I=-2) = 0.5*F_X(\frac{y}{2}) + 0.5*(1 - F_X(\frac{y}{2}))$$

Correct me if I am wrong. Any comments are appreciated.

Aaron Zeng
Aaron Zeng
October 20, 2014 03:56 AM

I correct Aaron Zeng's answer:

P(IX=2∣X=1) = P(IX=2,X=1)/ P(X=1) = P(I=2,X=1)/ P(X=1) = P(X=1|I=2)P(I=2)/ P(X=1) = 0.5P(X=1|I=2)/P(X=1)

P(Y=2) = P(IX=2) = P(IX=2,I=2) + P(IX=2,I=-2) = P(X=1) + P(X=-1) = 2P(X=1) (incase of mean μ = 0)

that shows P(IX=2∣X=1) # P(Y=2), so Y and X are dependent.

Nhan Nguyen
Nhan Nguyen
May 16, 2019 02:46 AM

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