Chapter 2 Section 1 Question 1

Question

Let $S=\{1,2,3...\}$ and let $X(s)=s^2$ and $Y(s)=\frac{1}{s}$ for $s\in S$. For each of the following quantities, determine (with explaination) whether or not it exists. If it does exist, then give its value.

- $min_{s\in S}X(s)$
- $max_{s\in S}X(s)$
- $min_{s\in S}Y(s)$
- $max_{s\in S}Y(s)$

Answer

Author: Mohammad-Ali Bandzar| Date:Oct 13 2020

- (a)$min_{s\in S}X(s)$
- Since $s^2$ is strictly increasing as $s \rightarrow \infty$ we know that X(1) will be our minimum value, Therefore $min_{s\in S}X(s)=1^2=1$
- (b)$max_{s\in S}X(s)$
- Since $s^2$ is strictly increasing without bound as $s \rightarrow \infty$ and since our sample space goes to positive infinity, we can conclude that X(s) is an unbounded random variable. Similar to Example 2.1.10 from the textbook. Therefore,$max_{s\in S}X(s)=DNE(Does Not Exist)$
- (c)$min_{s\in S}Y(s)$
- Does not exist, since our sample space S goes from $1\rightarrow\infty$

we should look at the function $f(x)=\frac{1}{x}$ where $x\in \mathbb{W}$ since our function is strictly decreasing on this domain naturally the largest number in our sample space S would produce our minimum value. Since we cannot plug $s=\infty$ into $Y(s)$ we know that there is no minimum.

In other words: since $\lim_{x\rightarrow\infty}f(x)=0$ our minimum value should be zero, but since there is no $s\in S$ such that $Y(s)=0$ the minimum does not exist. - (d)$max_{s\in S}Y(s)$
- Since our function is strictly decreasing on its domain(the sample space). We know that $Y(s)=max_{s\in S}Y(s)$ when s is the lowest possible number, in this case that means that $s=1$ therefore, $max_{s\in S}Y(s)=Y(1)=1$