independence of X^2 if we know X and Y are independent?

[oofiedoofie]

A. Let A and B be 2 independent random variables that are positive and have a variance that is not zero. Prove that [imath]A^2[/imath] and [imath]B[/imath] are independent.

B. Use the result from to show that the random variables W = A+ B and Z = AB
are positively correlated (i.e. Cov(W, Z) > 0)

What I know so far:

if ?⊥B then ??[?=?∩?=?]=??[?=?]∗??[?=?]. Since we know that A is positive then all the values in f(A) must be positive. How can I use this knowledge to prove that ??[?^2=?∩?=?]=??[?^2=?]∗??[?=?]? I tried to figure out how to determine the probability ?^2 or ??[?^2=?] in terms of ?. I've been stuck on this and don't know how to proceed.

 

[BigBeachBanana]

To prove that [imath]A^2[/imath] and [imath]B[/imath] are independent, we need to show that the joint probability distribution of [imath]A^2[/imath] and [imath]B[/imath] can be factored into the product of their individual probability distributions.

Let [imath]f_A(x)[/imath] and [imath]f_B(y)[/imath] be the probability density functions of A and B, respectively. Since A and B are independent, the joint probability density function of A and B is given by: [imath]f_{A,B}(x,y) = f_A(x) f_B(y)[/imath]

Now, let [imath]Z = A^2.[/imath] We want to find the probability density function of Z and B, denoted by [imath]f_{Z, B}(z,y).[/imath]

Show that [imath]f_{Z,B}(z,y) = f_{Z}(z) \times f_B(y)[/imath] using change of variables.


Side note: use [imath] [/imath] tags or \͏( and \͏) tags for inline typesetting. For other options, see the LaTeX thread on the News board.

 

[oofiedoofie]

Hi, thanks for the response but right now I only know elementary probability theory, so I don't know what this means exactly. I wanted to use an approach where we utilize the fact that probability of X is positive and non zero variance and also the definition of independence- if [imath] A \perp B[/imath] then [imath] Pr[A=a \cap B=b]= Pr[A=a] *Pr[B=b] [/imath].

 

[BigBeachBanana]

Hi, thanks for the response but right now I only know elementary probability theory, so I don't know what this means exactly. I wanted to use an approach where we utilize the fact that probability of X is positive and non zero variance and also the definition of independence- if [imath] A \perp B[/imath] then [imath] Pr[A=a \cap B=b]= Pr[A=a] *Pr[B=b] [/imath].

Are A and B discrete or continuous random variables? I'm asking because you're using the "=" e.g. [imath]\Pr(A=a)[/imath]. This is 0 if A and B are continuous random variables.

By definition,
[math]\Pr(A \le a \cap B \le b)= \Pr(A \le a) \times \Pr(B \le b)[/math]
We want to show
[math]\Pr(Z \le z \cap B \le b)= \Pr(Z \le z) \times \Pr(B \le b)[/math]
Since [imath]Z = A^2 \implies A = \sqrt{Z} \text{ for } A>0[/imath]

[math]\begin{aligned} P(Z \le z ,B\le b) &=P(A^2 \leq z, B \leq b) \\ &= P(A \leq \sqrt{z}, B \leq b) \\ &= P(A \leq \sqrt{z})P(B \leq b) \quad \because A \perp B \\ &= P(A^2 \leq z)P(B \leq b)\\ &= P(Z \leq z)P(B \leq b) \end{aligned}[/math]