Mean & Variance of the Exponential

What this lecture covers

This recitation problem works through the exponential distribution in detail: deriving its CDF from its PDF, computing its mean and variance using integration by parts, and then extending to the maximum and minimum of several independent, identically distributed exponential random variables. It builds on the definitions of PDF, CDF, expectation and variance from earlier in the course.

You'll come away able to derive the exponential's CDF, mean and variance from its density, and use the trick of rewriting max <= z and min > w as intersections of simpler events to find the distribution of extreme values of independent random variables.

Key ideas

• Exponential CDF: for X exponential with rate lambda, F(x) = 1 - e^(-lambda*x) for x >= 0, found by directly integrating the density.
• Mean via integration by parts: E[X] = 1/lambda, computed by choosing u=t and dv as the density term, with the exponential term dominating the polynomial term in the limit as t -> infinity.
• Variance from E[X^2]: a similar integration-by-parts calculation gives E[X^2] = 2/lambda^2, so Var(X) = 1/lambda^2.
• Exponentials always beat polynomials in the limit: any term like t * e^(-lambda*t) or t^2 * e^(-lambda*t) goes to 0 as t goes to infinity, which is why these integrals converge.
• Max of independent variables: P(max(X1,...,Xn) <= z) equals the product of the individual CDFs, since the max is below z exactly when every variable is below z.
• Min of independent variables: P(min(X1,...,Xn) > w) equals the product of the individual survival probabilities P(Xi > w), since the min exceeds w exactly when every variable exceeds w.
• Minimum of exponentials is exponential: the minimum of two independent exponentials with rate lambda is itself exponential, with rate 2*lambda.

Before you watch

• Know the exponential distribution's PDF and the definitions of CDF, expectation and variance.
• Be comfortable with integration by parts and with taking limits of products like t * e^(-t) as t -> infinity.
• This problem assumes you've seen continuous random variables and independence from earlier lectures in the course.

Check your understanding

1. Why does the derivation choose u=t rather than u = e^(-lambda*t) when applying integration by parts for the mean?
2. Why does P(max(X1,X2,X3) <= z) factor into a product of individual CDFs, but P(min(X1,X2) <= w) cannot be handled the same way directly?
3. Why is the minimum of two independent rate-lambda exponentials itself exponential, and what is its rate?
4. How would the formula for the PDF of the maximum change with four independent exponentials instead of three?