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Suppose $X_t$ and $Y_t$ are independent Poisson processes with parameters $\lambda_1$ and $\lambda_2$, respectively, measuring the number of calls arriving at two different phones. Let $Z_t=X_t+Y_t$.

(a) Show that $Z_t$ is a Poisson process. What is the rate parameter for $Z$? (b) What is the probability that the first call comes on the first phone? (c) Let $T$ denote the first time that at least one call has come from each of the two phones. Find the density and distribution function of the random variable $T$.

I checked $P(Zt+\Delta t =Zt+k)$ for $k=0$, $k=1$ and $k>1$ for the case (a) and it look's okay I think. On the other hand, I couldn't solve (b) and (c). How can I solve the parts (b) and (c)?

Ahaan S. Rungta
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Ninja
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2 Answers2

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Hint on c):

$$T>t\iff X_tY_t=0$$

drhab
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  • If I understood the question correctly, this is false. You want at least one of each type, so it is possible that $Z_t>0$ and $T<t$, if there have only been calls of one type. – QQQ Dec 14 '14 at 13:19
  • @Liron You are right. Thank you for correcting me. I repaired. – drhab Dec 14 '14 at 13:28
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First of all, look up the term "Superposition of a Poisson process".

a) The probability you wrote does not make sense, I guess some parenthesis is missing. First you need to know that the sum of any two independent Poisson random variables $X$ and $Y$ with rates, $\lambda_1$ and $\lambda_2$ is Poisson with the sum of rates. You can show this by computing the lst (or mgf),

$$Ee^{-\alpha(X+Y)}=Ee^{-\alpha X}Ee^{-\alpha Y}=e^{(\lambda_1+\lambda_2)(e^{-\alpha}-1)}.$$

Next you can use this to show that the increments of $Z_t$ are indeed independent and Poisson distributed.

b) Denote the first two calls of each type by $U_1\sim exp(\lambda_1)$ and $V_1\sim exp(\lambda_2)$. You need to compute the probability $P(U_1<V_1)$. This can be done using the properties of the Exponential distribution and simple integration. You should get: $$ P(U_1<V_1)=\frac{\lambda_1}{\lambda_1+\lambda_2}.$$

c) You wish to find the distribution of $T=U_1+V_1$. This is a convolution of two independent exponential random variables which can be found by conditioning: $$P(T<t)=P(U_1+V_1<t)=\int_{0}^t f_{U_1}(u)P(V_1<t-u)du.$$

QQQ
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