Expectation and Variance using Moment Generating Functions

Question

Use moment generating functions to verify the following:
The expected value of the sum of independent random variables is the sum of the expected values.

Initially I thought to use the property that if $X$ and $Y$ are independent, then $M_{X+Y}(s)=M_X(s)M_Y(s)$

This property states that the sum of independent variables can be computed using the individual product of their moment generating function.

Using the property above $$M_{X+Y}(s)=\int _{-\infty}^{\infty}\int _{-\infty}^{\infty}e^{s(x+y)}f_{X,Y}(x,y)dxdy$$ $$=\int _{-\infty}^{\infty}\int _{-\infty}^{\infty}e^{sx+sy}f_{X,Y}(x,y)dxdy$$ Not sure if this step is correct, since the two r.v. are independent their joint density can be expressed as their marginal i.e. $$=\int _{-\infty}^{\infty}\int _{-\infty}^{\infty}e^{sx+sy}f_X(x)f_Y(y)dxdy$$ Now we can derive $$=\frac{d}{ds}|s=0\int _{-\infty}^{\infty}\int _{-\infty}^{\infty}e^{sx+sy}f_X(x)f_Y(y)dxdy$$ $$=\int _{-\infty}^{\infty}\int _{-\infty}^{\infty}(x+y)f_X(x)f_Y(y)dxdy$$

Not sure if I am splitting this integral correctly, $$=\int _{-\infty}^{\infty}xf_X(x)dx+ \int _{-\infty}^{\infty}yf_Y(y)dy$$ $$=E(X)+E(Y)$$

I suppose if this is correct, I can do the same on of the RHS of the equation.

Answer 1

Hint: In addition to using the property (due to independence of $X$ and $Y$)

$$M_{X+Y}(s) = M_{X}(s) M_{Y}(s) $$

which you point out, also use the property of MGF (the reason behind the name moment "generating" function) that successive derivatives evaluated at $s=0$ give the successive moments. In particular, consider taking the first derivative of the above expression on both sides (right side needs product rule), evaluate them at 0, and then match.

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EDIT (May 2nd): It's perhaps best to work out the simple details instead of convoluted answers. This problem is a nice simple illustration of the properties of moment generating functions and as such the beauty shouldn't get lost going back to the definition/sums/integrals/etc. So, here we go:

Derivative of the left side $M'_{X+Y}(0)$ is $E(X+Y)$. Taking the derivative on the right side using the product rule and evaluating at $s=0$, we get:

$$M'_{X}(0) \times M_{Y}(0) \, + \, M_{X}(0) \times M'_{Y}(0) = EX \times 1 + 1 \times EY = EX + EY$$

since $M_{X}(0)=1$ and $M'_{X}(0)=EX$ for any moment generating function.

QED.

Answer 2

Given that the random variables $X$ and $Y$ are independent. Then, by the definition of MGF, we have, \begin{eqnarray*} M_{X+Y}(t)&=&E\left(e^{t(X+Y)}\right)=E(e^{tX})\cdot E(e^{tY})\qquad\text{(by independence)}\\ &=&E\left[ 1 + tX + \dfrac{t^2 X^2}{2!} + \cdots\right]\cdot E\left[ 1 + tY + \dfrac{t^2 Y^2}{2!} + \cdots\right]\\ &=&E\left[ 1 + tX + \dfrac{t^2 }{2!}X^2 + \cdots \right.\\ & &\quad\qquad tY +t^2 XY + \dfrac{t^3}{2!}X^2 Y + \cdots \\ & &\left.\quad\qquad \dfrac{t^2 Y^2}{2!} + \dfrac{t^3}{2!}XY^2 +\cdots \right] \\ M_{X+Y}^{'}(t)|_{t=0}&=& E(X+Y) = E(X)+E(Y) \end{eqnarray*}