Question

Exercise 1. C What is the expectation of X in Example 2 if the number of days until the virus is stopped is Geometric with expectation of 4 days?

Answer 1

1.

T ~ Geom(p = 1/4)

The PMF of T is,

Based the on the example 2 C, the expectation of X is,

$E[X] = \sum_{t \in N}^{ }4000 * 1.25^tP(T = t) =\sum_{t \in N}^{ }4000 * \left ( \frac{5}{4} \right )^t * \left ( \frac{3}{4} \right )^{t-1} \frac{1}{4}$

$=\frac{5}{4} \sum_{t \in N}^{ }1000 * \left ( \frac{5}{4} \right )^{t-1} * \left ( \frac{3}{4} \right )^{t-1} = 1250 \sum_{t \in N}^{ } \left ( \frac{15}{16} \right )^{t-1}$

$= 1250 \frac{1}{1 - (15/16)}$ (Using Sum of Infinite Geometric series)

= 1250 * 16

= 20000