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Two independent observations $X_1$ and $X_2$ are generated from the Poisson distribution with mean 1. The experiment is said to be successful if $X_1 + X_2$ is odd. What is the expected value of the number of experiments required to obtain the first success? Help to solve this problem.

StubbornAtom
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Max
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    What have you tried? Can you, say, compute the probability that $X_1$ is even? – lulu May 17 '20 at 11:43
  • May it be of help that the sum is Poisson with mean $2$? – Michael Hoppe May 17 '20 at 12:02
  • Probability of even/odd results from a $\mathrm{Poisson}(\lambda)$ may be found here: https://math.stackexchange.com/questions/2007238/what-is-the-probability-of-getting-an-even-number-from-a-poisson-random-draw. – Minus One-Twelfth May 17 '20 at 13:09
  • i can find distribution of $X1+X2$ but for odd I don't know how to find – Max May 17 '20 at 16:03

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You can actually solve it with an MC approach. The starting state, $\emptyset$ and the abosrbent state is $S_{odd/even, even/odd}$. This is because the only way to get an odd sum is for $X_1$ to be even and $X_2$ odd or vice versa. The probability of this event is $$ p_{\emptyset, 1} = P(X_1 = 2k+1)\times P(X_2=2k)\times 2 =2 e^{-2} \sum_{k \in \mathbb{Z} }\frac{1}{(2k+1)!} \times \sum_{k\in \mathbb{Z}}\frac{1}{(2k!)} $$ Now, to get the mean time until the state you need to compute $$ m_{\emptyset ,1 } = 1+ p_{\emptyset, 1} \times 0 + (1-p_{\emptyset, 1})\times m_{\emptyset ,1 } $$

Can you handle from here?

Alex
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