Expected Value at Infinity

$\lim_{P\to\infty} EV(P) = \lim_{P\to\infty} \sum_{k=1}^P \left(\frac{k}{P}\right)^2 \times \left(\frac{1}{P}\right) = \int_0^1 x^2\,dx = \frac{1}{3}$

The first element is $\frac{1}{P}$ , the second element is $\frac{2}{P}$ , the third element is $\frac{3}{P}$ , and so forth until the last element, which is $\frac{P}{P}$ .

Since there are $P$ elements, the probability of each element being selected is $\frac{1}{P}$ .

The possible results of an element being chosen at random and squared are $(\frac{1}{P})^2$ , $(\frac{2}{P})^2$ , $(\frac{3}{P})^2$ $...$ $(\frac{P}{P})^2$ .

Therefore, $EV(P)=\frac{1}{P} \times (\frac{1}{P})^2 + \frac{1}{P} \times (\frac{2}{P})^2 + \frac{1}{P} \times (\frac{3}{P})^2... \frac{1}{P} \times (\frac{P}{P})^2$ .

We can factor out $\frac{1}{P} \times (\frac{1}{P})^2$ , leaving us with $EV(P)=\frac{1}{P^3}\times(1^2+2^2+3^2...P^2)$ .

The series $1^2+2^2+3^2...P^2$ can be represented using the formula $\frac{P(P+1)(2P+1)}{6}$ . Plugging that into our equation gives $EV(P)=\frac{1}{P^3} \times \frac{P(P+1)(2P+1)}{6}$ .

We can simplify:

$EV(P)=\frac{(P+1)(2P+1)}{6P^2}$

$EV(P)=\frac{2P^2+3P+1}{6P^2}$

$EV(P)=\frac{1}{6} \times (\frac{2P^2}{P^2}+\frac{3P}{P^2}+\frac{1}{P^2})$

$EV(P)=\frac{1}{6} \times (2+\frac{3}{P}+\frac{1}{P^2})$

Now, we can plug in $\infty$ for $P$ . This makes $\frac{3}{P}$ and $\frac{1}{P^2}$ equal to 0, leaving us with $EV(P)=\frac{1}{6} \times 2$ , which simplifies to $EV(P)=\boxed{\frac{1}{3}}$ .

$EV(P)=\text{average of squares of elements}=\frac{\sum_{i=1}^P \left(\frac{i}{P}\right)^2}{P}=\frac{1}{P^3}\sum_{i=1}^P i^2$

$\sum_{i=1}^P i^2$ will be a cubic function of $P$ because the difference between consecutive sums $\sum_{i=1}^x i^2-\sum_{i=1}^{x-1} i^2=x^2$ is quadratic. We can therefore say $\sum_{i=1}^P i^2=aP^3+bP^2+cP+d$ . Noting that $\sum_{i=1}^P i^2=\sum_{i=0}^P i^2$ , we have the following system of equations:

$\begin{cases} d & =0\\ a+b+c & =1\\ 8a+4b+2c & =1+4=5\\ 27a+9b+3c & =5+9=14 \end{cases}$

Solving, we get

$\begin{bmatrix} a\\ b\\ c \end{bmatrix}=\text{rref}\left(\begin{bmatrix} 1 & 1 & 1 & 1\\ 8 & 4 & 2 & 5\\ 27 & 9 & 3 & 14 \end{bmatrix}\right) \times \begin{bmatrix} 0\\ 0\\ 0\\ 1 \end{bmatrix}=\begin{bmatrix} 1/3\\ 1/2\\ 1/6 \end{bmatrix}$

Therefore,

$EV(P)=\frac{\frac{P^3}{3}+\frac{P^2}{2}+\frac{P}{6}}{P^3}=\frac{1}{3}+\frac{1}{2P}+\frac{1}{6P^2}\\ \\ \lim_{P\rightarrow \infty} EV(P)=\boxed{\frac{1}{3}}$