Simplification of variance of a conditional distribution expression

Question

Suppose we have a collection of 8 independently and identically distributed exponential random variables $X_i \stackrel{\text{iid}}{\sim} \text{Exp}(1)$ and we select $N$ of them, where $N$ ranges from $1$ to $8$ with equally likely probabilities. We are interested in determining the variance of the minimum of the selected $N$.

It is given that the minimum of a group of exponentially distributed variables $X_i$ is given by $$\min\{x_i\} \sim \text{Exp}\left(\sum_{\forall i} \lambda_i\right)$$, so the conditional distribution of the minimums, say, $T$, is given by $$T \mid N = \text{Exp}(N).$$

Using the formula $$\text{Var}(T) = \mathbb{E}[\text{Var}(T \mid N)] + \text{Var}(\mathbb{E}[T \mid N])$$ yields an expression that I have difficulty simplifying: \begin{align*} \text{Var}[T] &= \mathbb{E}[\text{Var}[T \mid X]] + \text{Var}[\mathbb{E}[T \mid X]] \\ &= \mathbb{E}[\text{Var}[\text{Exp}(N)]] + \text{Var}[\mathbb{E}[\text{Exp}(N)]] \\ &= \mathbb{E}[1/N^2] + \text{Var}[1/N] \\ &= \sum_{i = 1}^{8} \frac{1}{i^2}P_N(i) + \sum_{i = 1}^{8} \left(\frac{1}{i} - \mathbb{E}[1/N]\right)^2P_N(i) \\ &= \frac{1}{8}\left(\sum_{i = 1}^{8} \frac{1}{i^2} + \sum_{i = 1}^{8} \left(\frac{1}{i} - \frac{1}{8}\sum_{i = 1}^{8} \frac{1}{i}\right)^2 \right) \end{align*} Any advice on how to proceed?

Answer 1

What you are looking at are the first and second order harmonic numbers.

They do not have nice closed forms.

If you do not have a table or such handy, well you have a series of eight fractions. Just add them up manually, or by computation.

$${\sum_{i=1}^{8}\tfrac 1i =H_8=\tfrac{761}{280} \:,\\\sum_{i=1}^{8}\tfrac 1{i^2}=H_8^{(2)}=\tfrac{1077749}{705600}}$$

Answer 2

The hierarchical model is $$T \mid N \sim \operatorname{Exponential}(n), \quad f_{T \mid N}(t) = n e^{-n t}, \quad t > 0, \\ N \sim \operatorname{DiscreteUniform}(8), \quad \Pr[N = n] = 1/8, \quad n \in \{1, 2, \ldots, 8\}.$$

The conditional expectation and variance are $$\operatorname{E}[T \mid N] = 1/N, \\ \operatorname{Var}[T \mid N] = 1/N^2.$$ The law of total variance then gives $$\operatorname{Var}[T] = \operatorname{E}[\operatorname{Var}[T \mid N]] + \operatorname{Var}[\operatorname{E}[T \mid N]] = \operatorname{E}[1/N^2] + \operatorname{Var}[1/N].$$ The first term is easy enough to compute, but the second requires an additional step: $$\operatorname{Var}[1/N] = \operatorname{E}[1/N^2] - \operatorname{E}[1/N]^2.$$ So $$\begin{align} \operatorname{Var}[T] &= 2 \operatorname{E}[1/N^2] - \operatorname{E}[1/N]^2 \\ &= \frac{1}{4} \sum_{n=1}^8 \frac{1}{n^2} - \left(\frac{1}{8} \sum_{n=1}^8 \frac{1}{n}\right)^2 \\ &= \frac{1}{4} \frac{1077749}{705600} - \left( \frac{1}{8}\cdot \frac{761}{280} \right)^2 \\ &= \frac{2406379}{9031680}. \end{align}$$ Unfortunately, there just isn't any simpler way to evaluate this expression.