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I have two Poisson distributions with parameters $\lambda 1$ and $\lambda 2$.
Now I want to find the average of minimum value of this two distributions.

I mean that if I derive two random numbers from these distributions and find the minimum of this two values, and repeat this experiment so many times, what would be the average of the minimum value.

Is this true to consider min ($\lambda 1$ , $\lambda2$ ) as the minimum value?

I appreciate your help. It is very important to me.

Thanks

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1 Answer 1

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For $i=1,2$ let $X_i$ be independent random variables Poisson-$\lambda_i$ distributed.

You seem to be asking:

"Is it true that in this case $\mathbb E\min(X_1,X_2)=\min(\mathbb EX_1,\mathbb EX_2)$?"

The answer is: "no".

It is evident that $\min(X_1,X_2)\leq X_1$ and $P(\min(X_1,X_2)<X_1)>0$.

These observations justify the conclusion that $\mathbb E\min(X_1,X_2)<\mathbb EX_1=\lambda_1$.

Likewise we find that $\mathbb E\min(X_1,X_2)<\mathbb EX_2=\lambda_2$.

This together implies that $$\mathbb E\min(X_1,X_2)<\min(\lambda_1,\lambda_2)$$

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  • $\begingroup$ Thank you for you answer. So how can I find the expected value of minimum values? $\endgroup$
    – Taban
    Commented May 10, 2015 at 12:28
  • $\begingroup$ I haven't thought about it yet. Actually I gave an answer to the question in the body and not the question in the title. $\endgroup$
    – drhab
    Commented May 10, 2015 at 12:30
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    $\begingroup$ You could give this a try: $\mathbb{E}\min\left(X_{1},X_{2}\right)=\sum_{n=0}^{\infty}P\left(\min\left(X_{1},X_{2}\right)>n\right)=\sum_{n=0}^{\infty}P\left(X_{1}>n\wedge X_{2}>n\right)=\sum_{n=0}^{\infty}P\left(X_{1}>n\right)P\left(X_{2}>n\right)$. Succes not garanteed. Good luck. $\endgroup$
    – drhab
    Commented May 10, 2015 at 12:44

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