Let $X_i\sim\text{Pois}(\lambda_i)$ for $i=1,2,\ldots,n$ and $Y = \min X_i$. Can we show that, for example $\mathbb{E}[Y] \leq f(\lambda,n)\min\lambda_i$ for some $f : (\mathbb{R}^n,\mathbb{N}) \to [0,1]$ and lower bound the variance $\mathbb{V}[Y]$ by anything meaningful?

Jensen's inequality tells us $$\mathbb{E}[Y] = \mathbb{E}[\min_i X_i] \leq \min_i\mathbb{E}[X_i] = \min_i\lambda_i$$ already but I'd like something more concrete.

Note that this question asks a similar question, but is concerned with the entire distribution over $Y$ while I only need bounds on 2 moments. Accordingly, I'd expect a more closed-form answer available.

If it helps, in my case $X_i\sim\text{Pois}(\lambda)$ for $i=1,\ldots,n-1$ and $X_n \sim \text{Pois}(\lambda + \gamma)$ for $\lambda,\gamma>0$ so for $\gamma$ and $n$ large enough we may effectively assume $X_i\sim\text{Pois}(\lambda)$ are i.i.d (with high probability) for the purposes of estimating $Y$.

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    $\begingroup$ What do you mean by "assume $X_i$" in the last sentence? Are some words missing? $\endgroup$
    – whuber
    Jan 20, 2018 at 17:41
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    $\begingroup$ @whuber ah thanks. I meant that $X_i$ are effectively i.i.d because for large enough $\gamma$, the probability that $Y = X_n$ is low. $\endgroup$ Jan 20, 2018 at 18:02
  • $\begingroup$ are you sure you can use Jensen's inequality here? $\endgroup$
    – Taylor
    Jan 21, 2018 at 5:55
  • $\begingroup$ @Taylor I’m quite sure: the minimum of a collection of linear functionals over $\mathbb{R}^n$ is concave (draw a picture to check). Denote $X=(X_1,\ldots,X_n)\in\mathbb{R}^n$. Then $\min X_i = \min \langle X,e_i\rangle$ is a minimum of linear functionals and hence concave, so Jensen applies. $\endgroup$ Jan 21, 2018 at 6:41
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    $\begingroup$ You do not even need Jensen: $\min X_i\le X_i$ for all $i$'s, hence $\mathbb{E}[\min_i X_i] \leq\mathbb{E}[X_i]$ for all $i$'s. $\endgroup$
    – Xi'an
    Jan 21, 2018 at 10:01

1 Answer 1


This only answers half of my question. Lower bounding the variance of $Y$ is still open.

Without loss of generality assume $\lambda_{\min} = \lambda_1\leq \lambda_2\leq \dots \leq \lambda_n = \lambda_{\max}$. Note that $$ \mathbb{P}(Y > 0) = \prod_{i=1}^n (1 - e^{-\lambda_i}). $$ Moreover, $\mathbb{E}[Y | Y>0] \leq \mathbb{E}[X_1 | X_1>0]$. Bringing these together,

\begin{align} \mathbb{E}[Y] &= \mathbb{E}[Y | Y>0]\mathbb{P}(Y > 0)\\ &\leq \mathbb{E}[X_1 | X_1>0]\mathbb{P}(Y>0)\\ &= \lambda_{\min} \frac{\prod_{i=1}^n(1 - e^{-\lambda_{i}})}{1 - e^{-\lambda_{\min}}}\\ &= \lambda_{\min} \prod_{i=2}^n (1 - e^{-\lambda_i})\tag{$*$}\\ &\leq \lambda_{\min} (1 - e^{-\lambda_{\max}})^{n-1} \end{align} as desired. Note that this is sharp for $n=1$, and depending on the situation $(*)$ might be more helpful than the final bound.


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