Doubt

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Doubt

Halflife

While calculating the variance of Z, are the two terms XY and X/Y independent? If yes then how?
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Re: Doubt

Abhitesh
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.
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Re: Doubt

Halflife
Thanks.
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Re: Doubt

Sachin Sehgal
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.
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Re: Doubt

Abhitesh
X|Y is the value of X given Y.
The value of X doesn't depend on Y.
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Re: Doubt

Halflife
I don't think it's 'X given Y'. It seems to be 'X/Y' only.
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Re: Doubt

Abhitesh
In that case they would be equal.
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Re: Doubt

AdityaGarg13
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!!!
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Re: Doubt

Halflife
In reply to this post by Abhitesh
Thanks.
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Re: Doubt

Halflife
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.
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Re: Doubt

sneha
only if I assume X to be independent ....I'm able to solve the question

otherwise this is getting complicated
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Re: Doubt

Halflife
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.
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Re: Doubt

sneha
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)
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Re: Doubt

Halflife
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.
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Re: Doubt

Shree
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..:-)
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Re: Doubt

Halflife
We both have solved it taking example distributions only. You just made the distribution of y simpler with y=1 with probability 1. So your method is better. Thanks.
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