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

oofiedoofie

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A. Let A and B be 2 independent random variables that are positive and have a variance that is not zero. Prove that A2A^2 and BB 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.
 
To prove that A2A^2 and BB are independent, we need to show that the joint probability distribution of A2A^2 and BB can be factored into the product of their individual probability distributions.

Let fA(x)f_A(x) and fB(y)f_B(y) 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: fA,B(x,y)=fA(x)fB(y)f_{A,B}(x,y) = f_A(x) f_B(y)

Now, let Z=A2.Z = A^2. We want to find the probability density function of Z and B, denoted by fZ,B(z,y).f_{Z, B}(z,y).

Show that fZ,B(z,y)=fZ(z)×fB(y)f_{Z,B}(z,y) = f_{Z}(z) \times f_B(y) 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.
 
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 AB A \perp B then Pr[A=aB=b]=Pr[A=a]Pr[B=b] Pr[A=a \cap B=b]= Pr[A=a] *Pr[B=b] .
 
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 AB A \perp B then Pr[A=aB=b]=Pr[A=a]Pr[B=b] Pr[A=a \cap B=b]= Pr[A=a] *Pr[B=b] .
Are A and B discrete or continuous random variables? I'm asking because you're using the "=" e.g. Pr(A=a)\Pr(A=a). This is 0 if A and B are continuous random variables.

By definition,
Pr(AaBb)=Pr(Aa)×Pr(Bb)\Pr(A \le a \cap B \le b)= \Pr(A \le a) \times \Pr(B \le b)
We want to show
Pr(ZzBb)=Pr(Zz)×Pr(Bb)\Pr(Z \le z \cap B \le b)= \Pr(Z \le z) \times \Pr(B \le b)
Since Z=A2    A=Z for A>0Z = A^2 \implies A = \sqrt{Z} \text{ for } A>0

P(Zz,Bb)=P(A2z,Bb)=P(Az,Bb)=P(Az)P(Bb)AB=P(A2z)P(Bb)=P(Zz)P(Bb)\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}
 
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