I'm dealing with the exponential distribution with mu$\neq$ 0 and am trying to find some expectations. I'm stumped. can anyone take a look at it for me?

here's my question.

$x_1,...x_n$ are iid with this pdf $\sigma^{-1}e^{-(x-\mu)/\sigma}I(x>\mu)$ with both parameters unknown.


I'm trying to find the distribution of $U$, $x_{n:1}$ is the first order statistic

I changed U up a little as

$$\sum(x_i-x_{n:1})=x_1-x_{n:1}+x_2-x_{n:1}+...+x_n-x_{n:1}=\sum x_i-nx_{n:1}$$ then i just have to find the distribution of $x_i$ and use convolution to find the distribution of the pair of terms.

my problem is finding the distribution of sum of xi's. I'm not sure how to do it. I tried adding two together and then generalizing the answer, but it didn't work.

$$V=X_1+X_2$$ $$\int_\mu^v \sigma^{-1}e^{-(x_1-\mu)/\sigma}I(x_1>\mu)\sigma^{-1}e^{-((v-x_1)-\mu)/\sigma}I((v-x_1)>\mu)=$$

$$\sigma^{-2}ve^{-(v-2\mu)/\sigma}-\sigma^{-2}\mu e^{-(v-2\mu)/\sigma}$$

does anyone know how to find the distribution of $U$?

  • 1
    $\begingroup$ This question seems to qualify for the self-study tag, please add it. $\endgroup$
    – Xi'an
    Feb 18, 2016 at 9:37

1 Answer 1


You don't need to find the entire distribution, just get the expected value directly.

First, this particular distribution is that of an exponential$(\sigma)$ random variable conditioned on being greater than $\mu$, which you can verify by direct calculation of the density function. This is nice as the exponential distribution is memoryless, which means the amount by which it exceeds any given quantity conditioned on it being greater than that quantity has exactly the same distribution as the original random variable.

Now if we look at $U$ notice that only $n - 1$ terms in the sum are nonzero, and the nonzero terms all have the same mean, so

$$ \text{E}(U) = (n - 1) \text{E}(X_{j^\star} - X_{(1)}) $$

where $j^\star$ is the index of any term that is not the minimum, and $X_{(1)}$ is the sample minimum.

But $X_{j^\star} - X_{(1)}$ is nothing but the amount by which $X_{j^\star}$ exceeds $X_{(1)}$ given that it's larger than $X_{(1)}$ which we already know is distributed exponential$(\sigma)$ due to memorylessness (the added conditioning on being larger than $\mu$ becomes irrelevant). This means the expectation above is just $\sigma$ and so $\text{E}(U) = (n - 1) \sigma$.

  • $\begingroup$ Why are $X_i-X_{n:1}$ and $X_j-X_{n:1}$ independent random variables? (the identically distributed part is OK). And why would the answer change at all if we don't assume independence? After all, linearity of expectation does not need independence, and that's all that it appears you would be needing. $\endgroup$ Feb 17, 2016 at 20:46
  • $\begingroup$ @DilipSarwate I'm fairly sure they are, but I don't know how to prove it off the top of my head. I can edit that part out if in fact all the poster wants to know is the expected value. $\endgroup$
    – dsaxton
    Feb 17, 2016 at 20:53

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