Given that $X_1,\dots, X_n$ are independent random variables. Suppose $M = \min(X_1, X_2,\dots, X_n)$ and $X_i$ are exponential random variables with parameter $λ_i$, compute $E[M X_j | M = X_i]$ where $i \ne j$.

I have got the pdf and pmf of $M$. But I still don't know how to solve. Anyone can help?

  • 2
    $\begingroup$ Assuming that $M=X_i$, $M\not =X_j$. ($i\not = j$) Let alone that $E[M=X_j|...]$ is an undefined notation. $\endgroup$
    – zoli
    Commented Apr 18, 2015 at 12:29
  • $\begingroup$ Sorry.. That is a typo. I've revised it. $\endgroup$
    – Euclid Ye
    Commented Apr 18, 2015 at 12:46
  • 1
    $\begingroup$ > OP writes: "I have got the pdf and pmf of M". /// What is the "pdf AND pmf of M"?? $\endgroup$
    – wolfies
    Commented Apr 18, 2015 at 17:46
  • $\begingroup$ What is the intent of the conditioning that the sample minimum $M = X_i$, and what do you intend it to mean? Because as it stands, I suspect this is a question that some well-meaning person is going to go to a lot of trouble to solve, only to find out that you intended something quite different. $\endgroup$
    – wolfies
    Commented Apr 18, 2015 at 18:06

2 Answers 2


We have $$\mathbb{E}\left[MX_j\mid M=X_i\right] =\frac{\mathbb{E}\left[MX_j:M=X_i\right]}{\mathbb{P}\left[M=X_i\right]}$$ since $\mathbb{P}\left[M=X_i\right]>0$.

We note $\Lambda=\prod_{k\neq i}\lambda_k$, $\lambda=\sum_{1\leq k\leq n}\lambda_k$ and $\alpha=\sum_{k\neq i,j}\lambda_k$ ; we have $$\mathbb{E}\left[MX_j:M=X_i\right] =\mathbb{E}\left[X_iX_j\mathfrak{1}_{M=X_i}\right] =\mathbb{E}\left[\mathbb{E}\left[X_iX_j\mathfrak{1}_{M=X_i}\mid X_i\right]\right]$$ with $$\mathbb{E}\left[X_iX_j\mathfrak{1}_{M=X_i}\mid X_i\right] =\Lambda\int_{\left(\mathbb{R}_+\right)^{n-1}}X_ix_j \mathfrak{1}_{M=X_i} e^{-\lambda_{1}x_1}\ldots e^{-\lambda_{i-1}x_{i-1}}e^{-\lambda_{i+1}x_{i+1}}\ldots e^{-\lambda_{n}x_n}\mathrm{d}x_{1}\ldots \mathrm{d}x_{i-1}\mathrm{d}x_{i+1}\ldots \mathrm{d}x_{n}$$ $$=\lambda_jX_i e^{-\alpha X_i}\int_{X_i\leq x_j}x_je^{-\lambda_j x_j}{d}x_{j} =\lambda_jX_i e^{-\alpha X_i}\frac{e^{-\lambda_j X_i}}{\lambda_j^2}\left(1+\lambda_jX_i\right) =\frac{1}{\lambda_j}X_i\left(1+\lambda_jX_i\right)e^{-\left(\alpha+\lambda_j\right) X_i}$$

so that

$$\mathbb{E}\left[MX_j:M=X_i\right] =\frac{1}{\lambda_j}\mathbb{E}\left[X_i\left(1+\lambda_jX_i\right)e^{-\left(\alpha+\lambda_j\right) X_i}\right] =\frac{1}{\lambda_j}\int_{x_i\geq0}x_i\left(1+\lambda_jx_i\right)e^{-\left(\alpha+\lambda_j\right) x_i}\lambda_i e^{-\lambda_i x_i}\mathrm{d}x_i$$ $$=\frac{\lambda_i}{\lambda_j}\int_{x_i\geq0}x_i\left(1+\lambda_jx_i\right)e^{-\lambda x_i} \mathrm{d}x_i =\frac{\lambda_i}{\lambda_j}\frac{\lambda+2\lambda_j}{\lambda^3}.$$

Next, we work on the term $\mathbb{P}\left[M=X_i\right]$. We have :

$$\mathbb{P}\left[M=X_i\right] =\mathbb{P}\left[X_i\leq X_1,\ldots X_i\leq X_n\right] = \mathbb{E}\left[\mathbb{P}\left[X_i\leq X_1,\ldots X_i\leq X_n\mid X_1,\ldots,X_{i-1},X_{i+1},\ldots,X_n\right]\right]$$ $$=\mathbb{E}\left[\mathbb{P}\left[X_i\leq\min_{\substack{1\leq k\leq n\\k\neq i}}X_k\right]\right] =1-\mathbb{E}\left[e^{-\lambda_i\min_{k\neq i}X_k}\right]$$

where we used the fact that $X_i$ is independant of $\sigma\left(X_k;k\neq i\right)$. Then, we compute

$$\mathbb{E}\left[e^{-\lambda_i\min_{k\neq i}X_k}\right] =\int_{\left(\mathbb{R_+}\right)^{n-1}}e^{-\lambda_i\min_{k\neq i}x_k}\Lambda e^{-\lambda_1 x_1}\ldots e^{-\lambda_{i-1}x_{i-1}}e^{-\lambda_{i+1}x_{i+1}}\ldots e^{-\lambda_1 x_1} \mathrm{d}x_{1}\ldots \mathrm{d}x_{i-1}\mathrm{d}x_{i+1}\ldots \mathrm{d}x_{n}$$

$$=\sum_{\substack{\ell=1\\\ell\neq i}}^{n}\Lambda\int_{\left(\mathbb{R_+}\right)^{n-1}}e^{-\lambda_ix_{\ell}} e^{-\lambda_1 x_1}\ldots e^{-\lambda_{i-1}x_{i-1}}e^{-\lambda_{i+1}x_{i+1}}\ldots e^{-\lambda_1 x_1} \mathfrak{1}_{x_{\ell}\leq x_1,\ldots x_n}\mathrm{d}x_{1}\ldots \mathrm{d}x_{i-1}\mathrm{d}x_{i+1}\ldots \mathrm{d}x_{n}$$

$$=\sum_{\substack{\ell=1\\\ell\neq i}}^{n}\lambda_{\ell}\int_{x_{\ell}\geq0}e^{-\left(\lambda_1+\ldots+\lambda_n\right)x_{\ell}}\mathrm{d}x_{\ell} =\frac{1}{\lambda}\sum_{\substack{\ell=1\\\ell\neq i}}^{n}\lambda_{\ell} =\frac{\lambda-\lambda_{i}}{\lambda} =1-\frac{\lambda_{i}}{\lambda}.$$

Finally, we obtain

$$\mathbb{E}\left[MX_j\mid M=X_i\right] =\frac{\lambda_i}{\lambda_j}\frac{\lambda+2\lambda_j}{\lambda^3}\frac{\lambda}{\lambda_i} =\frac{\lambda+2\lambda_j}{\lambda_j\lambda^2}.$$

  • $\begingroup$ "Now by independ(e)nce, we have $\mathbb{E}\left[MX_j:M=X_i\right] =\mathbb{E}\left[X_iX_j\right]$" Hmmm... Why exactly? $\endgroup$
    – Did
    Commented Apr 18, 2015 at 17:30
  • $\begingroup$ The independance was used here to write $\mathbb{E}\left[X_iX_j\right] =\mathbb{E}\left[X_i\right]\mathbb{E}\left[X_j\right]$. EDIT: but there is a problem indeed : I wrote that $MX_j\mathfrak{1}_{M=X_i} =X_iX_j$, and it is not true... $\endgroup$
    – Nicolas
    Commented Apr 18, 2015 at 17:34
  • $\begingroup$ Yes, $MX_j1_{M=X_i}=X_iX_j1_{M=X_i}$, not $X_iX_j$. $\endgroup$
    – Did
    Commented Apr 18, 2015 at 17:37
  • $\begingroup$ Agree! I am working on the numerator :-) $\endgroup$
    – Nicolas
    Commented Apr 18, 2015 at 17:40
  • $\begingroup$ Edited. Hope there is no problem in the calculus. I think all is ok now... $\endgroup$
    – Nicolas
    Commented Apr 18, 2015 at 18:22

This is not an answer!

This is just a long comment to Nicolas' answer.

I doubt that the following is true

$$\mathbb{P}\left[M=X_i\right] =\mathbb{P}\left[X_i\leq X_1,\ldots X_i\leq X_n\right]=$$

$$\frac{\lambda_1+\ldots+\lambda_{i-1}+\lambda_{i+1}+\ldots+\lambda_n}{\lambda_1+\ldots+\lambda_n}.$$ The expression above says that if $\lambda_i=0$ then $$\mathbb{P}\left[M=X_i\right]=1.$$ In words: the smaller $\lambda_i$, the higher the probability is that $X_i$ is the minimum.

If we consider,$X_{\lambda}$, an exponentially distributed random variable with parameter $\lambda$ then

$$\lim_{\lambda \rightarrow 0}\mathbb P(X_{\lambda}>t)=\lim_{\lambda \rightarrow 0}e^{-\lambda t}=1,$$ and this is true independendently of the size of $t$. To me, this tells that the smaller the $\lambda$ the smaller the probability is that $X_{\lambda}$ is small. This contradicts intuitively to the result quoted above.

So I did my own calculation for $\mathbb{P}\left[M=X_i\right].$

Here it is: $$\mathbb P(X_i\le X_1,X_i\le X_2, \cdots X_i\le X_n)=$$ $$=\prod_{k=1}^n \lambda_k\iint \cdots \int_{\{x_i\le x_1,x_i\le x_2, \cdots, x_i\le x_n\}}e^{-\sum_{k=1}^{n}\lambda_kx_k} \ \ dx_1dx_2 \cdots dx_n=$$ $$=\prod_{k=1}^n \lambda_k\int_0^{\infty}\left[\int_{x_i}^{\infty}\int_{x_i}^{\infty}\cdots \int_{x_i}^{\infty}e^{-\sum_{k=1}^{n}\lambda_kx_k}dx_1dx_2\cdots dx_{i-1}dx_{i+1}\cdots dx_{n}\right]dx_i=$$ $$=\prod_{k=2}^n\lambda_k\int_0^{\infty}e^{-\lambda_1x_i}\left[\int_{x_i}^{\infty}\int_{x_i}^{\infty}\cdots \int_{x_i}^{\infty}e^{-\sum_{k=2}^{n}\lambda_kx_k}dx_2\cdots dx_{i-1}dx_{i+1}\cdots dx_{n}\right]dx_i=$$ $$=\lambda_i\int_0^{\infty}e^{-x_i\sum_{k=1}^n\lambda_k}dx_i=\frac{\lambda_i}{\sum_{k=1}^n\lambda_k}.$$

  • $\begingroup$ I agree with your answer : I wrote the complementary probability in my former answer that explains that I got $1-$ "your result". $\endgroup$
    – Nicolas
    Commented Apr 19, 2015 at 9:56

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