# Doubt

16 messages
Open this post in threaded view
|
Report Content as Inappropriate

## Doubt

 While calculating the variance of Z, are the two terms XY and X/Y independent? If yes then how?
Open this post in threaded view
|
Report Content as Inappropriate

## Re: Doubt

 XY = X if Y=1      = -X if Y=-1. X/Y = X for any Y. Now the value of XY will depend on X/Y. If X/Y = x then XY = x or -x with equal probability.
Open this post in threaded view
|
Report Content as Inappropriate

## Re: Doubt

 Thanks.
Open this post in threaded view
|
Report Content as Inappropriate

## Re: Doubt

 In reply to this post by Abhitesh But why X/Y is X ..I mean it's value should also be x or -x depending upon the value of Y.
Open this post in threaded view
|
Report Content as Inappropriate

## Re: Doubt

 X|Y is the value of X given Y. The value of X doesn't depend on Y.
Open this post in threaded view
|
Report Content as Inappropriate

## Re: Doubt

 I don't think it's 'X given Y'. It seems to be 'X/Y' only.
Open this post in threaded view
|
Report Content as Inappropriate

## Re: Doubt

 In that case they would be equal.
Open this post in threaded view
|
Report Content as Inappropriate

## Re: Doubt

 In reply to this post by Halflife I too think so. If it was X given by, it would have been enclosed in bracket. Other term is product, this one would be division!!!
Open this post in threaded view
|
Report Content as Inappropriate

## Re: Doubt

 In reply to this post by Abhitesh Thanks.
Open this post in threaded view
|
Report Content as Inappropriate

## Re: Doubt

 In reply to this post by AdityaGarg13 We need not go so far. I only needed to know if the covariance term is negative or positive that is enough for this question. Thank you btw.
Open this post in threaded view
|
Report Content as Inappropriate

## Re: Doubt

 only if I assume X to be independent ....I'm able to solve the question otherwise this is getting complicated
Open this post in threaded view
|
Report Content as Inappropriate

## Re: Doubt

 Take the following distributions : X : -2, 0, 2 with P(X) : 1/3, 1/3, 1/3 & Y : -1, 1 with P(Y) : 1/2, 1/2. Now V(Z) = V(xy + x/y) = V(xy) + V(x/y) + Cov(xy, x/y) Now form distributions of xy and x/y and calculate their variances from that. As for the covariance, it's coming as : E((xy.x/y)^2) - (E(xy.x/y))^2   {standard formula of covariance). It'll become : E((x^2)) - (E(x))^2 ie it's the variance of x only. Both terms positive. Therefore : greater.
Open this post in threaded view
|
Report Content as Inappropriate

## Re: Doubt

 hey,as per what I read cov(x,y)=E(XY)-E(X)E(Y) and acc to this it will become E(XY.X/Y)-E(XY)*E(X/Y)
Open this post in threaded view
|
Report Content as Inappropriate

## Re: Doubt

 Oh sorry I got it wrong by mistake. Yeah your formula is right but by that too the covariance is coming 8/3 (for the example distributions that I've taken) as the E(xy) term is 0.
Open this post in threaded view
|
Report Content as Inappropriate

## Re: Doubt

 In reply to this post by Halflife Cant it be solved like this:   Let y=1 Z=X+X    =2X V(Z)=4V(X) Let y=-1 Z=-X-X    =-2X V(Z)=4V(X) And in both the cases  V(Z)>V(X) If this is not the correct way then please explain why it isnt..:-)