# Thread: Expected Value of normally distributed random variables

1. ## Expected Value of normally distributed random variables

Given two independent random variables Y and W. Where Y is distributed as N(0,1) and W is distributed as N(0.100).

a) Show that E[Y] = 1 and E[W^2) = 100
b) show that E[Y^3] = 0 and E[W^3] = 0.
c) Show that E[Y^4] = 3 and E[W^4] = 3*100^2

I thought that if Y~N(0,1) then E[Y] would equal 0. I hope I don't need help with all of these, some guidance for part (a) would be much appreciated. Thanks.

2. ## Re: Expected Value of normally distributed random variables

E[Y] would equal 0. My guess is that fitting with the pattern of b) and c) that you're supposed to show that E[Y^2] = 1.

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alias (10-30-2011)

4. ## Re: Expected Value of normally distributed random variables

For the case where E[W^2] = 100, in (a), could I relate it to variance by:
var(W) = E[W-(E(W))]^2
= E[W^2] - (E[W])^2
= E[W^2] - 0, because E[W] = 0, so the E[W^2] is equal to the variance, which is 100. If this is right I can solve the rest in the same way.

5. ## Re: Expected Value of normally distributed random variables

That's correct.

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alias (10-30-2011)

7. ## Re: Expected Value of normally distributed random variables

I'm having trouble with part (c), showing that E[Y^4] = 3, after I expand and simplify E[(Y-E[Y])^4] I get an E[Y], which equals 0, in every term except E[Y^4] so my answer turns out as
E[Y^4] = 0.

8. ## Re: Expected Value of normally distributed random variables

Originally Posted by alias
I'm having trouble with part (c), showing that E[Y^4] = 3, after I expand and simplify E[(Y-E[Y])^4] I get an E[Y], which equals 0, in every term except E[Y^4] so my answer turns out as
E[Y^4] = 0.
You have to assume what the functional form of the pdf is to get the expectation you're trying to solve for i.e.

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alias (10-31-2011)

10. ## Re: Expected Value of normally distributed random variables

Or if you already know Stein's Lemma you could use that to simplify things a little bit. But really the problems at hand are very simplified special cases of uses of Stein's lemma so I doubt you've covered it already. Unless the point is to get you to use Stein's lemma...

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alias (10-31-2011)

12. ## Re: Expected Value of normally distributed random variables

I'm supposed to use these formulas for parts (b) and (c), for both variables Y and W:
b) E[(Y-μ)^3]/σ^3
c) E[(Y-μ)^4]/[E(Y-μ)^2]^2
I've come up with nothing while trying to use either one of the equations and I am especially confused as to how I to E(Y^4) = 3 for part (c). I am not supposed to use Stein's Lemma, not sure I would know how to.

13. ## Re: Expected Value of normally distributed random variables

You could go the direct route as Dragan suggests and do the integration (hint: use integration by parts).

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alias (11-01-2011)

15. ## Re: Expected Value of normally distributed random variables

what would you use for [u, du, dv, v] to integrate by parts? I know it works but I have to show all the steps of integration.

16. ## Re: Expected Value of normally distributed random variables

In my last post I understated my integration ability. I have ABSOLUTELY NO IDEA how to integrate that function.

17. ## Re: Expected Value of normally distributed random variables

consider... . Now since you want to reduce the problem to an easier one that you've previously solved you'll probably want to take the derivative of the x^3 and integrate the other part (which hopefully you can do... if not see the spoiler for a hint).
Spoiler:

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alias (11-02-2011)

19. ## Re: Expected Value of normally distributed random variables

There are several approach for the even normal moments; the direct way, as pointed out above, is to consider the integral as the gamma integral.

One other possible way is to use the moment generating function of normal.

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alias (11-02-2011)

21. ## Re: Expected Value of normally distributed random variables

This is true. alias have you covered moment generating functions? That would definitely be an alternative way to do this that doesn't require quite as much thought if you already know what the MGF of the normal is. Although deriving the MGF just to use it in this particular case would probably be overkill and a little bit more work than just doing it directly. I'm not saying it's particularly hard but there is some completing the square to be had and a little bit of tedious bookkeeping to keep track of.

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alias (11-02-2011)

23. ## Re: Expected Value of normally distributed random variables

I haven't covered mgf's yet. There not in my econometrics text. I've tried to find some material on them but what I've found so far I don't unserstand.

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