Expectation of Normal Distribution

[MattFox]

Let Y = a + bZ + cZ2 where Z (0,1) is a standard normal random variable.
(i) Compute E[Y], E[Z], E[YZ], E[Y^2] and E[Z^2].
HINT: You will need to determine E[Z^r], r = 1, 2, 3, 4. When r = 1, 2 you should
use known results. Integration by parts will help when r = 3, 4.
I am struggling with the part of the question involving E[Z^3] and E[Z^4]. Clearly E[Z]=0 and E[Z^2]=1 but I do not no where to proceed when computing higher powers of Z. Any help would be greatly appreciated.

Thanks

 

[EnumaElish]

Expected value of a function g of z is E[g(z)] = $\int_{-\infty}^{+\infty}g(z)f(z)dz$ where f is the pdf of z.

 

[MattFox]

So am I right in thinking I let g(z) = z^3 and f(z)= 1/(sigma * sqrt2pi) ... and then use integration by parts?

 

[EnumaElish]

That should be it.

 

[MattFox]

I'm not really getting anywhere with the integration by parts. Is there any hint you could offer or show me a step I could be missing out?

 

[EnumaElish]

Can you post what you have so far?

 

[MattFox]

Ok so,

INT u dv = uv - INT v du

Then i let u = Z^3 and dv = the pdf of the normal dist.

Im lost when it comes to integrating the pdf and I'm not sure how it will condense down to a manageable solution. Sorry about the lack of symbols, I'm not sure how to properly upload images yet.

 

[EnumaElish]

If dv = the pdf of the normal dist., then what is v?

(If dv = f(z) and v = F(z), what is F?)

 

[MattFox]

F(z) is the cdf of the normal distribution.

So E(Z^3) = Z^3 * F(Z) - INT 3Z^2 * F(Z).

Do i then need to apply integration by parts again? I'm sorry if I'm asking too many questions I just really don't understand what I'm trying to achieve as a solution.

 

[MattFox]

Or more accurately, what form the solution will take i.e an integer, function of Z...

 

[EnumaElish]

u = Z^3 and dv = f(z)dz ===> v = F(z)

E(Z^3) = INT u dv = [uv] - INT v du = [Z^3 * F(Z)] - INT F(Z) 3Z^2 dZ, where each of the [uv] expression and the last INT are evaluated over (-infinity, +infinity).

E.g., [uv] = u(+infinity)v(+infinity) - u(-infinity)v(-infinity) ... but (infinity)^3 is not a real number, so something's amiss. See http://en.wikipedia.org/wiki/Integration_by_parts

 

[EnumaElish]

Another way to calculate E[z^n] is through the moment generating function. The normal dist. has moment generating function $M(t) = \exp\left(\mu t+\sigma^2 t^2/2\right)$, and its n'th moment is given by the n'th derivative of M w/r/t t, evaluated at t = 0.

See http://en.wikipedia.org/wiki/Normal_distribution
and http://en.wikipedia.org/wiki/Moment-generating_function