$X_1, \dots, X_n$ are independent random variables. Suppose $M = \min(X_1, X_2, \dots, X_n)$ 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? 
 A: 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
$$\mathbb{E}\left[MX_j\mid M=X_i\right]
 =\frac{\lambda_i}{\lambda_j}\frac{\lambda+2\lambda_j}{\lambda^3}\frac{\lambda}{\lambda_i}
 =\frac{\lambda+2\lambda_j}{\lambda_j\lambda^2}.$$
A: 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:
$$\mathbb P(X_i\le X_1,X_i\le X_2, \cdots X_i\le X_n)=$$
$$=\prod_{k=1}^n \lambda_k\iint \cdots \int_{\{x_i\le x_1,x_i\le x_2, \cdots, x_i\le x_n\}}e^{-\sum_{k=1}^{n}\lambda_kx_k} \ \ dx_1dx_2 \cdots dx_n=$$
$$=\prod_{k=1}^n \lambda_k\int_0^{\infty}\left[\int_{x_i}^{\infty}\int_{x_i}^{\infty}\cdots \int_{x_i}^{\infty}e^{-\sum_{k=1}^{n}\lambda_kx_k}dx_1dx_2\cdots dx_{i-1}dx_{i+1}\cdots dx_{n}\right]dx_i=$$
$$=\prod_{k=2}^n\lambda_k\int_0^{\infty}e^{-\lambda_1x_i}\left[\int_{x_i}^{\infty}\int_{x_i}^{\infty}\cdots \int_{x_i}^{\infty}e^{-\sum_{k=2}^{n}\lambda_kx_k}dx_2\cdots dx_{i-1}dx_{i+1}\cdots dx_{n}\right]dx_i=$$
$$=\lambda_i\int_0^{\infty}e^{-x_i\sum_{k=1}^n\lambda_k}dx_i=\frac{\lambda_i}{\sum_{k=1}^n\lambda_k}.$$
