# Did I mess up the Poisson-Gamma relationship?

Let $X_1, X_2$ be independent, exponentially distributed random variables with mean 2. So $X_1+X_2=Z$ is gamma distributed with $\alpha=2$ and $\beta=1/2$.

I am trying to solve the probability that $X_1>3$ given that $X_1 +X_2>3$. This should be fairly straightforward, yet my answer differs from the book's answer.

I set up the solution as $$P[X_1>3]/P[Z>3]=e^{-1.5}/(1-P[Z<3])=e^{-1.5}/(1-P[Q<2])$$ where $Q$ has a poisson distribution with mean equal to $x*\beta=3/2.$ As a result, I get $$e^{-1.5}/(1-[1.5e^{-1.5}+e^{-1.5}])=.223/.442 = .50.$$

My book's answer is $.40.$ Does anyone see where I might have messed up?

• You could integrate by parts for calculating the probability $P(X_1+X_2>3)$ from the density of $X_1+X_2$, which indeed gives $5e^{-3/2}/2$. Jul 17, 2018 at 19:32

\begin{aligned} P(X_1 > 3 \mid X_1 + X_2 > 3) & = \frac{P(X_1 + X_2 > 3 \mid X_1 > 3)P(X_1 > 3)}{P(X_1 + X_2 > 3)} \\ & = \frac{P(X_1 > 3)}{P(X_1 + X_2 > 3)} \text{ since X_2 > 0 with prob. 1} \\ & = \frac{e^{-1.5}}{P(X_1 + X_2 > 3)} \text{ using Exp(\frac{1}{2}) cdf} \\ \end{aligned}

You could easily evaluate $P(X_1 + X_2 > 3)$ by conditioning on $X_2$ and applying the law of total probability. But if you insist on thinking about it in terms of a Poisson process, you can do the following.

In a Poisson process with rate $\lambda$, the event where the sum of the first two inter-arrival times $X_1 + X_2$ is greater than 3 is precisely the event where 1 or fewer arrivals occurred in the time period up to 3.

Since $\lambda = \frac{1}{2}$, the number of arrivals from time 0 to 3, which we'll call $N(3)$, is distributed as Poisson($\frac{3}{2}$). Then, we have

$$P(N(3) \leq 1) = P(N(3) = 0) + P(N(3) = 1) = e^{-3/2} + \frac{3}{2}e^{-3/2} = \frac{5}{2}e^{-3/2}$$

and therefore

$$\frac{e^{-1.5}}{P(X_1 + X_2 > 3)} = \frac{e^{-3/2}}{\frac{5}{2}e^{-3/2}} = \frac{2}{5} = 0.4$$

• Because when $Z < 3 \iff Q \geq 2$, but you mistook it for $Z < 3 \iff Q < 2$, which is not the right interpretation (i.e. if two arrivals happened within 3 seconds, that means that the number of arrivals in 3 seconds is at least 2). Jul 17, 2018 at 21:18

Z denotes the point in time when the second Poisson event occurred. Z>3 means that the second Poisson event occurred after time 3, and therefore it is equivalent to Q<2 (Q being the number of Poisson events up to time 3) and not to Q>=2 as you calculated. If you divide 0.223 by 1-0.442=0.558, you will get the correct answer 0.4.

• But I directly substituted $P[Q<2]$ for $P[Z<3]$, so I'm not sure what you mean. Thank you for the response. Jul 17, 2018 at 19:14
• That's your mistake - P(Q<2) is the same as P(Z>=3) and not P(Z<3). Jul 18, 2018 at 2:54