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 $Y$.

  • 1
    $\begingroup$ What do you mean by "assume $X_i$" in the last sentence? Are some words missing? $\endgroup$
    – whuber
    Commented Jan 20, 2018 at 17:41
  • 1
    $\begingroup$ @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. $\endgroup$ Commented Jan 20, 2018 at 18:02
  • $\begingroup$ are you sure you can use Jensen's inequality here? $\endgroup$
    – Taylor
    Commented 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. $\endgroup$ Commented Jan 21, 2018 at 6:41
  • 1
    $\begingroup$ 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
    Commented 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.