1) For fX to be a pdf, the integral from -inf to +inf must =1. So you integrate fX = Ce^-(x^4) from -inf to +inf and set it equal to 1. Then solve for C.

2) Given Y=aX, then Var(Y) = a^2 * Var(X). But then you don't know Var(X)? A brute force way to find Var(X) = E(X^2) - E^2(X), and find the expectations by integrating x^2 * fX

I am taking my first probability course and need some help. I am working on some homework. The question I am on now asks...

Consider fX(x)=Ce^-(x^4)

Calculate the normalization factor C which makes fX a pdf.

thanks for the help on this folks, I found C= 2/[gamma(1/4)]

Calculate the constant a>0 for which Y=aX has a standard deviation = 1.

I found EX to be = gamma(3/2)/[2*gamma(1/4)]

EX^2= gamma(5/4)/[2*gamma(1/4)]

Var X= 1/8 - 1/64 = 7/64 (used numerical values for gamma fxns from wikipedia)

and used VarY=a^2 * VarX to solve for a which i found a= [8*sqrt(7)]/7

What is the pdf of the random variable Y?

therefore the pdf of Y must be 2/[gamma(1/4)] * e^(-(8^4/49)*y^4)

What is the fourth moment of Y?

I am getting some crazy stuff for this, using brute force to find EY,EY^2,EY^3, etc. This question seems to invite the use of the mgf in some way, but i am missing the trick to do that. Hmm maybe i should just generate the mgf with e^tx