# 14.4 - Special Expectations

14.4 - Special Expectations

The special expectations, such as the mean, variance, and moment generating function, for continuous random variables are just a straightforward extension of those of the discrete case. Again, all we need to do is replace the summations with integrals.

Expected Value

The expected value or mean of a continuous random variable $$X$$ is:

$$\mu=E(X)=\int^{+\infty}_{-\infty} xf(x)dx$$

Variance

The variance of a continuous random variable $$X$$ is:

$$\sigma^2=Var(X)=E[(X-\mu)^2]=\int^{+\infty}_{-\infty}(x-\mu)^2 f(x)dx$$

Alternatively, you can still use the shortcut formula for the variance, $$\sigma^2=E(X^2)-\mu^2$$, with:

$$E(X^2)=\int^{+\infty}_{-\infty} x^2 f(x)dx$$

Standard Deviation

The standard deviation of a continuous random variable $$X$$ is:

$$\sigma=\sqrt{Var(X)}$$

Moment Generating Function

The moment generating function of a continuous random variable $$X$$, if it exists, is:

$$M(t)=\int^{+\infty}_{-\infty} e^{tx}f(x)dx$$

for $$-h<t<h$$.

As before, differentiating the moment generating function provides us with a way of finding the mean:

$$E(X)=M'(0)$$

and the variance:

$$\text{Var}(X)=M^{\prime\prime}(0)-\left(M^\prime(0)\right)^2$$

## Example 14-2 Revisited Again

Suppose $$X$$ is a continuous random variable with the following probability density function:

$$f(x)=3x^2, \qquad 0<x<1$$

What is the mean of $$X$$?

#### Solution

What is the variance of $$X$$?

## Example 14-8

Suppose $$X$$ is a continuous random variable with the following probability density function:

$$f(x)=xe^{-x}$$

for $$0<x<\infty$$. Use the moment generating function $$M(t)$$ to find the mean of $$X$$.

#### Solution

The moment generating function is found by integrating:

$$M(t)=E(e^{tX})=\int^{+\infty}_0 e^{tx} (xe^{-x})dx=\int^{+\infty}_0 xe^{-x(1-t)}dx$$

Because the upper limit is $$\infty$$, we can rewrite the integral using a limit:

$$M(t)=\lim\limits_{b \to \infty} \int_0^b xe^{-x(1-t)}dx$$

Now, you might recall from your study of calculus that integrating this beast is going to require integration by parts. If you need to integrate

$$\int udv$$

integration by parts tells us that the integral is:

$$\int udv=uv-\int vdu$$

In our case, let's let:

$$u=x$$ and $$dv=e^{-x(1-t)}$$

Differentiating $$u$$ and integrating $$dv$$, we get:

$$du=dx$$ and $$v=-\dfrac{1}{1-t}e^{-x(1-t)}$$

Therefore, using the integration by parts formula, we get:

$$M(t)=\lim\limits_{b \to \infty} \left\{\left[-\dfrac{1}{1-t}xe^{-x(1-t)}\right]_{x=0}^{x=b}-\left(-\dfrac{1}{1-t}\right)\int_0^be^{-x(1-t)}dx\right\}$$

Evaluating the first term at $$x=0$$ and $$x=b$$, and integrating the last term, we get:

$$M(t)=\lim\limits_{b \to \infty}\left\{\left[-\dfrac{1}{1-t} be^{-b(1-t)}\right]+\left(\dfrac{1}{1-t}\right) \left[\left(-\dfrac{1}{1-t}\right)e^{-x(1-t)}\right]_{x=0}^{x=b} \right\}$$

which, upon evaluating the last term at $$x=0$$ and $$x=b$$, as well as simplifying and distributing the limit as $$b$$ goes to infinity, we get:

$$M(t)=\lim\limits_{b \to \infty}\left[-\dfrac{1}{1-t} \dfrac{b}{e^{b(1-t)}}\right]-\left(\dfrac{1}{1-t}\right)^2 \lim\limits_{b \to \infty}(e^{-b(1-t)}-1)$$

Now, taking the limit of the second term is straightforward:

$$\lim\limits_{b \to \infty}(e^{-b(1-t)}-1)=-1$$

Therefore:

$$M(t)=\lim\limits_{b \to \infty}\left[-\dfrac{1}{1-t} \dfrac{b}{e^{b(1-t)}}\right]+\left(\dfrac{1}{1-t}\right)^2$$

Now, if you take the limit of the first term as $$b$$ goes to infinity, you can see that we get infinity over infinity! You might recall that in this situation we need to use what is called L'HÃ´pital's Rule. It tells us that we can find the limit of that first term by first differentiating the numerator and denominator separately. Doing just that, we get:

$$M(t)=\lim\limits_{b \to \infty}\left[-\dfrac{1}{1-t} \times \dfrac{1}{(1-t)e^{b(1-t)}}\right]+\left(\dfrac{1}{1-t}\right)^2$$

Now, if you take the limit as $$b$$ goes to infinity, you see that the first term approaches 0. Therefore (finally):

$$M(t)=\left(\dfrac{1}{1-t}\right)^2$$

as long as $$t<1$$. Now, with the hard work behind us, using the m.g.f. to find the mean of $$X$$ is a straightforward exercise:

 [1] Link ↥ Has Tooltip/Popover Toggleable Visibility