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It is your birthday and you are waiting for someone to write a “Happy Birthday” message on your Facebook wall. Your waiting time is approximately Exponential with average waiting time of $10$ minutes between such postings; assume that the times of the postings are independent.

so $10 = E(X) = \cfrac{1}{\lambda} = \cfrac{1}{1/10}$ so $\lambda = \cfrac{1}{10}$

a. What is the probability that the next posting takes $15$ minutes or longer to appear?

$P(X \geq 15) = e^{-15/10} = .2231$

b. What is the standard deviation of the time in between consecutive Happy Birthday messages?

$Var(X) = \cfrac{1}{\lambda^2} = \cfrac{1}{1/100} = 100$ so $sd(X) = \sqrt{100} = 10$ minutes

c. Suppose that the most recent posting was done at 1:40 PM, and it is now 1:45 PM (i.e., no postings have been made during the last five minutes). What is the expected time for the next message to appear?

Since $E(X) = 10$ minutes per message and $5$ minutes have already passed, does this mean that the expected time for the next message to appear is $5$ minutes (from $10 - 5 = 5$ minutes)?

Can someone tell me if I understand part c) and have it right? (and also part a and b))?

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    Part c) is not correct! The exponential distribution is memoryless. – littleO Apr 05 '21 at 02:57
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    Analogously, suppose you toss a coin until it lands heads up. And suppose that so far you've tossed five tails in a row. What's the expected number of tosses until you get heads? The fact that you just tossed five tails in a row is irrelevant. You might say that the coin has no memory. – littleO Apr 05 '21 at 03:08
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    Oh so it's still 10 minutes cause of the memoryless property? No matter how long we wait? –  Apr 05 '21 at 03:21

1 Answers1

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Part $a$ is correct. You can check this by integrating the PDF ($1$ - area between $0$ and $15$ minutes):

$$1 - \int_0^{15} \frac{1}{10} e^{-1/10 x} \ dx = 1 - \frac{1}{10}\left[\frac{1}{-1/10} e^{-1/10 x}\right]_0^{15} = 1 + (e^{-15/10}-1) = e^{-15/10}$$

Part $b$ is also correct, but since $\frac{1}{\lambda} = 10$, the standard deviation can be obtained more directly from $\sqrt{(1/\lambda)^2}$.

Part $c$ is not correct, because the exponential distribution is memoryless. The proof for this property is sketched out here.

Toby Mak
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  • The geometric distribution can be seen as the discrete analogue of the exponential distribution. There is an interesting problem related to this one in this Maths SE thread. – Toby Mak Apr 05 '21 at 03:52