Computing expected cost for exponential random variable

Question

I am reading Probability and Statistics for Engineering and the Sciences.

Exercise 63, Chapter 4 says:

A consumer is trying to decide between two long-distance calling plans. The first one charges a flat rate of 10¢ per minute, whereas the second charges a flat rate of 99¢ for calls up to 20 minutes in duration and then 10¢ for each additional minute exceeding 20 (assume that calls lasting a noninteger number of minutes are charged proportionately to a whole-minute’s charge). Suppose the consumer’s distribution of call duration is exponential with parameter $\lambda$.

• Which plan is better if expected call duration is 10 minutes? 15 minutes?

Assuming the first question, when the duration is 10 minutes, I computed the cost of the first plan as:

$h_1(x) = 10 * E[x] = 10 * 10 = 100$

However, how do I compute the cost for the second plan ($h_2(x)$) ?

I tried with:

$h_2(x) = 99 * F(x \leq 20) + 10 * (1 - F(x \leq 20)) \approx 87$

But the correct result is $112.53$.

Answer 1

In the second plan, you pay $99$ cents for sure for a call, plus a charge for calls that last more than $20$ minutes. Let us compute the expected value of this additional charge.

The additional charge occurs with probability $e^{-20/10}$.

Given that a call lasted more than $20$ minutes, the expected additional time is $10$, by the memorylessness property of the exponential distribution. Thus the expected additional charge is $(10)(e^{-20/10})(10)$.

So the expected cost of a call on the second plan, when mean length is $10$, is $99+(10)(e^{-20/10})(10)$ cents.

The calculation when the mean is $15$ is done in the same way.

Answer 2

Let $X$ denote the call duration and $Y$ be the cost. We have $$ \mathbb{E}[Y \mid X \leq 20] = 99 \tag{$\clubsuit$} $$ and $$ \mathbb{E}[Y \mid X > 20] = 99 + 10 \cdot 10 = 199 \tag{$\spadesuit$} $$ The rationale for $(\spadesuit)$ is as follows. When $X > 20$, the call duration consists of two parts:

• The first $20$ minutes which costs totally $99$.

• The additional time which costs $10$ per minute. This time follows the same distribution of $X$ because exponential distribution is memoryless. Thus the expected cost is $10 \cdot \mathbb{E}[X] = 100$.

Therefore, $$ \mathbb{E}[Y] = \mathbb{E}[Y \mid X \leq 20] \cdot \Pr[X \leq 20] + \mathbb{E}[Y | X > 20] \cdot \Pr[X > 20] \approx 112.53 $$