How to calculate $E [Z^2]$ when $E[Z]$ is known?

For the following calculation, why $E[Z_1^2] = \pi/4$ ? In general, how to calculate $E[Z^2]$ ? Thanks much.

3 down vote

Let's also elaborate on Ross Millikan's answer, adapted to the case f(x)=1−x2−−−−−√, 0≤x≤1. Suppose that $(X1,Y1),(X2,Y2),\ldots$is a sequence of independent uniform vectors on $[0,1]\times [0,1]$, so that for each $i$, $X_i$ and $Y_i$ are independent uniform $[0,1]$ random variables. Define $Z_i$ as follows: $Z_i=1$ if $X_i^2+Y_i^2 \le 1$, $Z_i=0$ if $X_i^2+Y_i^2 > 1$, so the $Z_i$ are independent and identically distributed random variables, with mean $\mu$ given by $\mu = E[Z_1] =P[X_1^2+Y_1^2 \le 1]=P[(X_1,Y_1) \in \{(x,y) \in [0,1]^2 : x^2+y^2 \le 1\}] = \pi/4$, where the last equality follows from $P[(X1,Y1)\in A]=\text{area}A (A \subset [0,1]^2)$.

By the strong law of large numbers, the average Zˉn=∑ni=1Zin converges, with probability 1, to the expectation μ as n→∞. That is, with probability 1, Zˉn→π/4 as n→∞.

To get a probabilistic error bound, note first that the $Z_i$ have variance $\sigma^2$ given by $\sigma^2=\text{Var}[Z_1]=E[Z_1^2]−E^2[Z_1]=\pi /4−(\pi /4)^2=\pi /4 (1−\pi /4)<10/59$.

• Could you please edit your question to make it clearer? May 26, 2011 at 16:09

Elaborating on my (second, according to date) answer from Approximating $\pi$ using Monte Carlo integration.

Suppose that $(X_1,Y_1)$ is a uniform vector on $[0,1] \times [0,1]$, so that $X_1$ and $Y_1$ are independent uniform$[0,1]$ random variables. Define $Z_1$ as follows: $Z_1 = 1$ if $X_1^2 + Y_1^2 \leq 1$, $Z_1 = 0$ if $X_1^2 + Y_1^2 > 1$. Then, $${\rm E}[Z_1] = 1 \cdot {\rm P}[X_1^2 + Y_1^2 \leq 1] + 0 \cdot {\rm P}[X_1^2 + Y_1^2 > 1] = {\rm P}[X_1^2 + Y_1^2 \leq 1].$$ Similarly, $${\rm E}[Z_1^2] = 1^2 \cdot {\rm P}[X_1^2 + Y_1^2 \leq 1] + 0^2 \cdot {\rm P}[X_1^2 + Y_1^2 > 1] = {\rm P}[X_1^2 + Y_1^2 \leq 1].$$ Hence, ${\rm E}[Z_1^2] = {\rm E}[Z_1]$. Next, note that $X_1^2 + Y_1^2 \leq 1$ if and only if $(X_1,Y_1)$ belongs to the set $A \subset [0,1]^2$ defined by $$A = \lbrace (x,y) \in [0,1]^2 : x^2+y^2 \leq 1\rbrace.$$ Thus, $${\rm E}[Z_1^2] = {\rm P}[X_1^2 + Y_1^2 \leq 1] = {\rm P}\big[(X_1,Y_1) \in A].$$ However, ${\rm P}[(X_1,Y_1) \in A] = {\rm area}A$, and ${\rm area}A = \pi/4$ (recall that the area of a disc of radius $1$ is $\pi$); hence ${\rm E}[Z_1^2] = \pi/4$.

It is simply because Z square is equal to Z here. notice that Z square is equal to 1 iff Z is equal to 1 and Z square is equal to 0 iff Z is null.

An expectation operator is a basically integration. Provided you know the p.d.f. of a random variable $Z$, $f(z)$, $$E(Z)=\int z f(z) dz$$ and $$E(Z^{2})=\int z^2 f(z) dz$$ over the support (the interval of z in which $f(z)$ is positive) of $z$.

• Check your answer for typographical errors: you have two different definitions for $E(Z)$. Aug 6, 2012 at 2:58
• Sorry, but I don't see how this answer has any relevance to the question. Aug 6, 2012 at 14:49
• Sorry, I missreaded the question. I thought the question only goes in the first line. Nevertheless, the question is hard to understand anyway. He said "in general, how to calculate $E[Z^{2}]$." So I provided the solution to the question. Aug 7, 2012 at 2:00
• @Ikuyasu The account from which the question was posted does not exist anymore, so it is unlikely you will get any clarification (at least not from the original user). Aug 7, 2012 at 2:01
• It's pretty old post and somebody seemed to have gave a nice answer anyway. Aug 7, 2012 at 19:18