Proving Variance of a Random Variable with Moment-Generating Functions

[dmatador]

Suppose that Y is a random variable with moment-generating function m(t) and W = aY + b, with a moment-generating function of m(at) * e^(tb). Prove that V(W) = V(Y) * a^2. I have done an absurd amount of work on this problem, and I know its actual solution doesn't have one and a half pages worth of work needed. I have tried to find the variances separately and also to find the expected values, but this just gave me a big mess of equations and I need some advice.

 

[mathman]

E(W)= aE(Y) + b
E(W²)=a²E(Y²) + 2abE(Y) + b²
V(W)=E(W²)-(E(W))²=a²(E(Y²)-E(Y)²)=a²V(Y)

 

[dmatador]

Thanks a lot man. I kept trying to use the moment-generating function for W. This is a big help.

 

[statdad]

Are you sure you were not supposed to use the mgf?

$$\begin{align*}<br /> m_W(t) & = e^{tb} m_Y(at) \\<br /> m'_W(t) & = be^{tb} m_Y(at) + ae^{tb} m'_Y(at) \\<br /> \mu_W & = m'_W(0) = b + a\mu_Y <br /> \end{align*}$$

so the mean of W is $$a\mu_Y + b$$.

Remember that the second derivative, evaluated at t = 0, is [itex]\sigma^2 + \mu^2 [/tex].<br /> <br /> $$\begin{align*}<br /> m'_W(t) & = (bm_Y(at) + am'_Y(at)) e^{tb} \\<br /> m''_W(t) & =b(bm_Y(at) + am'_Y(at)) e^{b}+ (abm'_Y(at) + a^2m''_Y(at))e^{tb} \\<br /> m''_W(0) & = b(b + a\mu_Y) + (ab\mu_Y + a^2 (\sigma^2_y + \mu^2_Y)) \\<br /> & = a^2 \sigma_Y^2 + a^2\mu_Y^2 + 2ab\mu_Y + b^2 \\<br /> & = a^2 \sigma_Y^2 + (a\mu_Y + b)^2<br /> \end{align*}$$<br /> <br /> The final line in the second bit is $$E[W^2]$$, so the variance of W is<br /> <br /> $$a^2 \sigma^2_Y + (a\mu_Y+b)^2 - (a\mu_Y + b)^2 = a^2 \sigma^2_Y$$[/itex]