Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am given the following question:

Suppose that the times it takes for two students to solve a certain homework problem are independently and identically distributed according to the distribution $Poiss(\lambda)$.

Find the probability that one of the students will take at least twice as long as the other one to solve the problem.

What I did: Since $X,Y$ are independent $$P_{Y|X}(y|x)=P(Y=y|X=x)=P(Y=y)$$

Given some value, $k$, of $X$: The probability that it takes the second student at least twice as long to do the homework is $P(Y\geq2k)$.

Hence the probability that it takes the second student at least twice as long to do the homework is, according to Law of total probability, $$\sum_{k=1}^{\infty}P(X=k)P_{Y|X}(Y\geq2k|X=k)$$

$$=\sum_{k=1}^{\infty}P(X=k)\cdot P(Y\geq2k)$$




and this is where I am stuck.

Can someone please help me continue on calculating this sum, or maybe suggest a different approach ?

share|improve this question
Poisson is unusual assumption. Are you sure it is not exponential? And what about $k=0$? –  André Nicolas Dec 31 '12 at 16:29
@AndréNicolas - Yes, I have copied the question exactly how its written (and I to find it strange..). I am not sure about $k=0$, I assumed that it takes more than $0$ units of time to solve the homework –  Belgi Dec 31 '12 at 16:32
add comment

1 Answer

up vote 1 down vote accepted

The probability that the second student, $Y \sim \text{Poiss}(\lambda)$, independently takes at least twice as long as the first student, $X \sim \text{Poiss}(\lambda)$, to finish the test is given by:

$$\displaystyle \mathbb{P}(Y \ge 2 X) = \sum_{x = 0}^{\infty} \sum_{y = 2x}^{\infty} f_X(x) \frac{1}{2} f_Y \left( \frac{1}{2} y \right )$$

$f_Y(y) = 0$ for all $y \not \in \mathbb{N}_0$, so write the equivalent sum for $y = 2x + 2u$

$$\displaystyle = \frac{1}{2} \sum_{x = 0}^{\infty} \sum_{u = 0}^{\infty} f_X(x) f_Y (x + u )$$

This can be simplified further observing that this is an infinite sum of the cross-correlation $\left ( f_X \star f_Y \right )(u)$.

$$\displaystyle = \frac{1}{2} \sum_{u = 0}^{\infty} \left ( f_X \star f_Y \right )(u)$$

The cross-correlation of two Poisson distributions gives rise to the Skellam probability mass function $\displaystyle f_S(x) = e^{-(\lambda_1 + \lambda_2)} \left ( \frac{\lambda_1}{\lambda_2} \right ) ^ {x/2} I_{\lvert x \rvert}(2 \sqrt{\lambda_1 \lambda_2})$. Since the two random variables under consideration are i.i.d., this becomes an auto-correlation simplifying to $\displaystyle f_S(x) = e^{-2 \lambda } I_{\lvert x \rvert}(2 \lambda )$. (Observing the fact $\lambda > 0$ by definition.) Where $I_{x}(\cdot)$ is the modified Bessel function of the first kind.

$$\displaystyle = \frac{1}{2} e^{-2 \lambda } \sum_{u = 0}^{\infty} I_{\lvert u \rvert}(2 \lambda )$$

The infinite sum, $\displaystyle \sum_{k = 1}^{\infty} I_{k}(x)$ reduces to $\frac{1}{2} \left ( e^x - I_0(x) \right )$, thus:

$$\displaystyle \mathbb{P}(Y \ge 2 X) = \frac{1}{4} \left ( 1 + e^{-2 \lambda } I_0(2 \lvert \lambda \rvert) \right ) \quad \square$$

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.