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

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

$$=\sum_{k=1}^{\infty}P(X=k)\cdot(1-P(Y<2k))$$

$$=\sum_{k=1}^{\infty}P(X=k)\cdot(1-\sum_{j=1}^{2k-1}P(Y=j))$$

$$=\sum_{k=1}^{\infty}e^{-\lambda}\frac{\lambda^{k}}{k!}\cdot(1-\sum_{j=1}^{2k-1}e^{-\lambda}\frac{\lambda^{j}}{j!}))$$

and this is where I am stuck.

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

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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
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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:

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

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