Find $\lambda $ such that P(X=1)=$\frac{1}{2}$ where X is Poisson($\lambda$). Using the formula $P\left(X=x\right)=\frac{\lambda^xe^{-\lambda}}{x!}$ and plugging in 1 for x I was able to simplify it down to $\lambda e^{-\lambda}=\frac{1}{2}$. I was not too sure about how to solve this equation by hand, so I graphed the function of $\lambda e^{-\lambda}$ and y = $\frac{1}{2}$, but these graphs never intersect which makes it seem that there is never a time that P(X=1)=$\frac{1}{2}$ because the max of the graph of $\lambda e^{-\lambda}$ is at 0.368. Is this a correct assumption that there is no solution to this problem, if not, how would you solve this equation the proper way?
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3What you have said is correct, with the maximum being $e^{-1}\approx 0.368$ when $\lambda=1$. It is possible to have $P(X=0)=\frac12$ when $\lambda= \log_e(2) \approx 0.693$ but that would be a different question – Henry May 11 '19 at 23:55
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Your observation is correct and can be proved by differentiating $x e^{-x}$ and looking at the sign of the derivative : you find that the function is increasing on $]-\infty,1]$ and decreasing on $[1,+\infty[$. The maximum value is therefore $e^{-1}<\frac{1}{2}$. – Joel Cohen May 12 '19 at 00:01
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You are correct, there is no solution, and it’s impossible to achieve $1/2$. One intuitive reason is that $P(X=0)=P(X=1)$ at $\lambda=1$ (the maximizer) , which would imply $P(X>1)=0$, which would contradict the fact that the Poisson distribution takes on infinitely many values with positive probability.
It’s easy to show the maximizer is $\lambda=1$ by differentiating and then checking the second derivative at the critical point.
Alex R.
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