Minimize the Sum of Reciprocal of Probabilities

Question

I need a probability distribution over $ n $ events such that the sum of expected values is minimized. That is, minimize

The problem is given by:

$$\begin{aligned} \arg \min_{ {p}_{i} } & \; && \sum_{i = 1}^{n} \frac{1}{ {p}_{i} } \\ \text{subject to} & \; && \sum_{i = 1}^{n} {p}_{i} = 1 \end{aligned}$$

I guess it's a basic easy question. I would appreciate any hint.

Answer 1

Let's use the method of Lagrange multipliers:

$$ L=\sum_i\frac{1}{p_i}+\lambda\left(\sum_ip_i-1\right)\implies0=\frac{\partial L}{\partial p_i}=\lambda-\frac{1}{p_i^2}\implies\frac{\partial^2L}{\partial p_i^2}=\frac{2}{p_i^3}>0, $$

So it is guaranteed that the extreme point is a Minimum Point. Moreover:

$$ 0 = \lambda - \frac{1}{ {p}_{i}^{2} } \Rightarrow {p}_{i} = \frac{1}{ \sqrt{\lambda} } $$

And from the constraint:

$$ \sum_{i = 1}^{n} \frac{1}{ \sqrt{\lambda} } = 1 \Rightarrow \lambda = {n}^{2} $$

Which suggests that $ {p}_{i} = \frac{1}{n} $.

Answer 2

Here is an argument that tells you that the minimum value can only be attained the $p_i$'s are equal. Suppose the minimum value is attained when $p_i=q_i, 1 \leq i \leq n$. If possible let $q_i \neq q_j$. Note that $\frac 1 {q_i} +\frac 1 {q_j} >\frac 1 {\frac {q_i+q_j} 2}+\frac 1 {\frac {q_i+q_j} 2}$. [This is simply a re-writing of the inequality $(q_i-q_j)^{2} >0]$. Thus there is a choice of $p_i$'s for which we get a value lower than the minimum. This contradiction shows that $p_i$ are all equal when the minimum is attained.