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?

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

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].

$$\begin{align*} m'_W(t) & = (bm_Y(at) + am'_Y(at)) e^{tb} \\ m''_W(t) & =b(bm_Y(at) + am'_Y(at)) e^{b}+ (abm'_Y(at) + a^2m''_Y(at))e^{tb} \\ m''_W(0) & = b(b + a\mu_Y) + (ab\mu_Y + a^2 (\sigma^2_y + \mu^2_Y)) \\ & = a^2 \sigma_Y^2 + a^2\mu_Y^2 + 2ab\mu_Y + b^2 \\ & = a^2 \sigma_Y^2 + (a\mu_Y + b)^2 \end{align*}$$

The final line in the second bit is $$E[W^2]$$, so the variance of W is

$$a^2 \sigma^2_Y + (a\mu_Y+b)^2 - (a\mu_Y + b)^2 = a^2 \sigma^2_Y$$

 

[blue_raver22]

There are a few different approaches you can take to prove the variance of W is equal to the variance of Y multiplied by a^2. One approach is to use the definition of variance, which states that V(X) = E[(X - μ)^2], where μ is the mean or expected value of X.

Using this definition, we can start by finding the mean of W. Since W = aY + b, we can express the mean of W as E[W] = E[aY + b]. Using the linearity of expectation, we can rewrite this as E[W] = aE[Y] + b.

Next, we can find the variance of W by plugging in the definition of variance and the expression for the mean of W:
V(W) = E[(W - E[W])^2]
= E[(aY + b - (aE[Y] + b))^2]
= E[(aY - aE[Y])^2]
= a^2E[(Y - E[Y])^2]

Now, we can use the moment-generating functions of Y and W to simplify this expression. From the given information, we know that the moment-generating function of W is m(at) * e^(tb). Using the definition of moment-generating functions, we can write this as m(at) = E[e^(atY)].

Similarly, the moment-generating function of Y is m(t) = E[e^tY].

Using these two expressions, we can rewrite the variance of W as:
V(W) = a^2E[(Y - E[Y])^2]
= a^2E[e^(atY - atE[Y])]
= a^2E[e^(at(Y - E[Y]))]
= a^2m(at) * m(-at)

Since we know that m(t) * m(-t) = 1 (this is a property of moment-generating functions), we can simplify the above expression to:
V(W) = a^2m(at) * m(-at)
= a^2

Therefore, we have proven that V(W) = a^2, which is equal to V(Y) * a^2. This shows that the variance of W is equal to the variance of Y multiplied by a^2, as desired.

Another approach to this proof could be using the