Probability & Statistics Ch.2 Section 1 Question 1

Solutions for “Probability and Statistics: The Science of Uncertainty” (Second Edition). These are solutions I have come up with; I offer no guarantee of accuracy.

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 explanation) whether or not it exists. If it does exist, then give its value.

a) $\min_{s\in S}X(s)$

b) $\max_{s\in S}X(s)$

c) $\min_{s\in S}Y(s)$

d) $\max_{s\in S}Y(s)$

Solution

(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:

(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:

(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 $s=1$. Therefore: