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?

<br /> \begin{align*}<br /> m_W(t) &amp; = e^{tb} m_Y(at) \\<br /> m&#039;_W(t) &amp; = be^{tb} m_Y(at) + ae^{tb} m&#039;_Y(at) \\<br /> \mu_W &amp; = m&#039;_W(0) = b + a\mu_Y <br /> \end{align*}<br />

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

Remember that the second derivative, evaluated at t = 0, is \sigma^2 + \mu^2 [/tex].<br /> <br /> &lt;br /&gt; \begin{align*}&lt;br /&gt; m&amp;#039;_W(t) &amp;amp; = (bm_Y(at) + am&amp;#039;_Y(at)) e^{tb} \\&lt;br /&gt; m&amp;#039;&amp;#039;_W(t) &amp;amp; =b(bm_Y(at) + am&amp;#039;_Y(at)) e^{b}+ (abm&amp;#039;_Y(at) + a^2m&amp;#039;&amp;#039;_Y(at))e^{tb} \\&lt;br /&gt; m&amp;#039;&amp;#039;_W(0) &amp;amp; = b(b + a\mu_Y) + (ab\mu_Y + a^2 (\sigma^2_y + \mu^2_Y)) \\&lt;br /&gt; &amp;amp; = a^2 \sigma_Y^2 + a^2\mu_Y^2 + 2ab\mu_Y + b^2 \\&lt;br /&gt; &amp;amp; = a^2 \sigma_Y^2 + (a\mu_Y + b)^2&lt;br /&gt; \end{align*}&lt;br /&gt;<br /> <br /> The final line in the second bit is E[W^2], so the variance of W is<br /> <br /> &lt;br /&gt; a^2 \sigma^2_Y + (a\mu_Y+b)^2 - (a\mu_Y + b)^2 = a^2 \sigma^2_Y&lt;br /&gt;