Suppose there are $n=2k+1$ independent and identically distributed exponential random variables with rate parameter $\lambda.$ Find $\mathbb{E}[median(X_1,\cdots,X_n)]?$
To do so, I found the density of $median(X_1,\cdots,X_n) = X_{(k+1)},$ where $X_{(j)}$ denotes the jth order statistic. Namely,
$$f_{X_{(j)}}(a) = n \lambda e^{-\lambda a} \binom{n-1}{j-1}(1-e^{-\lambda a})^{(j-1)}(e^{-\lambda a})^{((n-1) -(j-1))} = n \lambda e^{-\lambda a} P(Y=j-1),
$$
where $Y$ is a binomial random variable with number of trials and probability of success equal to $n-1$ and $1-e^{-\lambda a},$ respectively. Hence,
$$f_{X_{(k+1)}}(a) = n \lambda e^{-\lambda a} \binom{n-1}{k}(1-e^{-\lambda a})^{k}(e^{-\lambda a})^{((n-1)-k)} = n \lambda e^{-\lambda a} P(Y=k).
$$ This leads to
$$
\mathbb{E}[median(X_1,\cdots,X_n)] = \mathbb{E}[X_{(k+1)}] = \int_{0}^{\infty} x f_{X_{(k+1)}}(x) dx,
$$
but I'm stuck on the integral– I'm not sure if there is a closed form solution.
Alternatively, is there a simpler way to find the solution?
Best Answer
Suppose $Y_1,\dotsc,Y_{m}$ are i.i.d exponential random variables with rate one. Consider the spacing variables $Z_{i} = Y_{(i)}-Y_{(i-1)}$ for $2\leq i \leq m$ where $Z_{1} = Y_{(1)}$. The joint density of the order statistics $(Y_{(1)}, \dotsc, Y_{(m)})$ is given by $$ f_{(Y_{(1)}, \dotsc, Y_{(m)})}(y_1,\dotsc,y_m) = m!f_{\mathbf{Y}}(y_1,\dots,y_m)=m!\exp\left(-\sum_{i=1}^{m} y_i\right) $$ where $y_{1}<y_2<\dotsb<y_{m}$. From this density and the mentioned transformation, one can deduce that the $Z_i$ are independent exponential random variables with rate $m-i+1$ for $i=1,\dotsc, m$. Because $$ Y_{(j)} =\sum_{i=1}^jZ_{i}, $$ it follows that $$ EY_{(j)} = \sum_{i=1} ^jEZ_{i} =\sum_{i=1}^j\frac{1}{m-i+1}.$$
In your particular case, $$ EX_{(k+1)}=\frac{1}{\lambda}\sum_{i=1}^{k+1}\frac{1}{2k+2-i}=\frac{1}{\lambda}(H_{2k+1}-H_{k}) $$ where $j=k+1$ and $m=2k+1$.