Density of a compound poisson process.

People arrive to a bank according to a possion process $N(t)$ with $\lambda = 1$ client/minute. Each client makes a deposit $Y \sim \mathrm{Unif}\{1,2\}$ in thousand dollars. Calculate the probability that at time $t=5$ there will be 6 thousand dollars.

The way I've trying to solve this is using compound poisson process:

Define $X(t)=\sum\limits_{i=1}^{N(t)}{Y_i}$ where $N(t)$ is the number of clients at time $t$ and $Y_i$ is the deposit of the client $i$.

Then, I need $\mathbb{P}[X(5)=6]$

I've been trying to find the density function like this:

$\mathbb{P}(X(t)=x)=\mathbb{P}\left(\sum\limits_{i=1}^{N(t)}{Y_i}=x\right)=\sum_{N(t)=0}^{\infty} \Bbb P\left( \sum_{i=1}^n Y_i=x|N(t)=n\right)\Bbb P(N(t)=n)$

if this is the right way to proceed, then I'm not sure how to keep going, so any help would be appreciated. Also, is there another way to get this problem solved?

• you forgot to multiply the conditional probability in the last step by $\Bbb P(N(t)=n)$ and then sum that expression over all possible values of $N(t)$ $\sum_{n=0}^{\infty} \Bbb P(\sum_{i=1}^n Y_i=x|N(t)=n)\Bbb P(N(t)=n)$ – user237393 May 5 '15 at 5:01
• columbia.edu/~ww2040/3106F14/lec1023.pdf for reference. – user237393 May 5 '15 at 5:02
• @user237393 The document only mentions means and variances of compound Poisson processes, while here it is a probability that is needed. One could use the normal approximation, but none of the quantities here are sufficiently large to justify its use. – alonso s May 5 '15 at 5:32
• @alonso s, it mentions the joint probability can be interpreted binomially. – user237393 May 5 '15 at 6:37

There must have been $3$, $4$, $5$, or $6$ clients in five minutes, for otherwise the total deposited could not be $6000$.

The number of clients has Poisson distribution with parameter $5$, so the probability of each of $3$, $4$, $5$, and $6$ are easily calculated. Call the probability there were $k$ clients $p(k)$.

For $3$ clients, all must have deposited $2000$, probability $\frac{1}{2^3}$.

For $4$ clients, there must have been two $2000$ and two $1000$. Probability is easily calculated, binomial. The probability is $\frac{\binom{4}{2}}{2^4}$.

For $5$ clients, again easy, probability of one head in $5$ tosses of a fair coin.

For $6$ clients, all must have deposited $1000$.

Finally, our probability is $p(3)\cdot\frac{1}{2^3}+p(4)\cdot \frac{\binom{4}{2}}{2^4}+\cdots$.

• The deposits are not either $1000$ or $2000$ with probability $1/2$ each, according to the problem description; they are uniformly distributed in the interval $[1000, 2000]$. – Brian Tung May 5 '15 at 6:05
• The problem says $\{1,2\}$ not $[1,2]$. And the probability of $6000$ with continuous distribution is $0$. – André Nicolas May 5 '15 at 6:09
• In that case, I read the problem as requiring a minimum of $6000$. But certainly your interpretation seems more tractable... – Brian Tung May 5 '15 at 6:11
• Too bad, I was hoping there would be an interesting solution to the continuous case (with the criterion being a minimum of $6000$). It leads to a reasonably convenient solution when the criterion is a minimum of $3000$. – Brian Tung May 5 '15 at 6:13
• Perhaps OP can clarify the intended meaning. One could interpret the question as meaning continuous uniform, and the $6000$ as meaning at least $6000$, though usually at this level people are careful to specify 'at least' when that is intended. Sums of $4$ or $5$ continuous uniforms are kind of messy. – André Nicolas May 5 '15 at 6:19