Variance calculation

[mkheidzedavid]

Hello Friends! I need urgent help. Got a midterm exam this evening.
So the problem is:

Assume that deltaS/S has normal distribution with mean of zero and variance sigma^2. Calculate variance of random variable (deltaS/S)^2.

Thank you very much

 

[ThinkDifferent]

use the fact that E[X^4] = 3, where X is standard normal.

 

[mkheidzedavid]

So it comes out that if we let sigmaS/S = X
Then
V(X^4) = E[(X^4-3)^2] = E(X^8-6X^4 +9) = Sum(x^8P(x)) - 6Sum(x^4P(x)) + 9

Wrong oops

 

[ThinkDifferent]

your question just asks:
Let Y be normal r.v. with mean = 0, variance = sigma^2. What is the variance of Y^2?

To answer, just represent Y as Y = sigma * X where X is standard normal and then use a definition of variance and the fact that E[X^4] = 3.

 

[AlexandreH]

Maybe:
E[(x-μ)^4]=3σ ^4
for C=0, E[x^4]=E[(x-μ)^4]=3σ^ 4

σ ^2= E[x^2]-μ^2, for μ=0 E[x^2]=σ ^2

Var[x^2]=E[(x^2 - E[x^2])^2]=E[(x^2 - σ ^2)^2]=E[x^4 + σ^4 -2x^2 σ^2]
= E[x^4]+ σ^4 -2 σ^2 σ^2
= 3σ^ 4 + σ^4 -2σ^4 = 2σ^4

What do you think?

even without having μ=0, this can be worked using E[(x-μ)^4]=3σ ^4 and E[(x-μ)^3]=0 and E[x^2]=σ ^2 + μ^2 and then expanding Var[x^2]=E[(x^2 - E[x^2])^2].

E[x^4]=μ4 + 6μ2σ2 + 3σ4

I ran this quickly and I think the solution is Var[x^2]=2σ^4 + 4μ2σ2

 

[AlexandreH]

E[(x-μ)^k]=0 for k odd
and
E[(x-μ)^k]=σ^k x (k -1) for k even

 

[bob]

Maybe:
E[(x-?)^4]=3??^4
for C=0, E[x^4]=E[(x-?)^4]=3?^?4

??^2= E[x^2]-?^2, for ?=0 E[x^2]=??^2

Var[x^2]=E[(x^2 - E[x^2])^2]=E[(x^2 - ??^2)^2]=E[x^4 + ?^4 -2x^2 ?^2]
= E[x^4]+ ?^4 -2 ?^2 ?^2
= 3?^?4 + ?^4 -2?^4 = 2?^4

What do you think?

even without having ?=0, this can be worked using E[(x-?)^4]=3??^4 and E[(x-?)^3]=0 and E[x^2]=??^2 + ?^2 and then expanding Var[x^2]=E[(x^2 - E[x^2])^2].

E[x^4]=?4 + 6?2?2 + 3?4

I ran this quickly and I think the solution is Var[x^2]=2?^4 + 4?2?2

Yes, that's right.

 

[bob]

E[(x-?)^k]=0 for k odd
and
E[(x-?)^k]=?^k x (k -1) for k even

That's not quite right. Take six derivatives of the MGF to see why.

 

[AlexandreH]

That's not quite right. Take six derivatives of the MGF to see why.

You are right, my bad, it worked for the first few items because 1!!=1 and 3!!=3...
it should be
E[(x-?)^k]=0 for k odd
and
E[(x-?)^k]=?^k x (k -1)!! for k even

where !! is double factorial.
X!!= product of all odd numbers from 1 to X included.
1!!=1
3!!=3
5!!=15
7!!=105
...