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Fix $\mu >0$ and take $\lambda \geq 0$. Let $B_p \sim \text{Ber}(p)$ with $p = \exp(-\mu - \frac{\lambda}2) $. Define the random variable $Y$ which is Poisson with parameter depending on the value of $B_p$: $$Y \overset{d}= (1 - B_p) \text{Poi}(\mu + \lambda) + B_p \text{Poi}(\mu + \frac \lambda 2)$$ all terms are independent. For any $\lambda$ I would like to find $\epsilon>0$ (possibly depending on $\lambda$) so that $Y$ stochastically dominates a $\text{Poi}(\lambda + \epsilon)$ random variable.

It is possible that this only works for $\mu$ sufficiently large.

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  • $\begingroup$ This isn't showing up when I compute the expectation. I'm getting $\mathbf {E} Y = -\frac 1 2 \lambda e^{-\mu - \lambda/2}+\lambda+\mu$. So we need $\frac 12 \lambda e^{- \lambda/2} < \mu e^\mu$, which holds for $\mu$ large enough. $\endgroup$
    – mathjunge
    Aug 23, 2014 at 23:49
  • $\begingroup$ Right, I messed up. Sorry about that. $\endgroup$
    – Did
    Aug 24, 2014 at 7:54

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