Consider $10000 \ N(0, 1)$-distributed r.v.’s and let $Z_n = \frac{1}{n} \sum_1^n X_i^2$
- Calculate $E(X^2_1 )$. How does this relate to the statement of the Law of Large Numbers?
- Explain how this can be used to give approximations for $E(g(X))$ for functions $g$. For which functions $g$ can this be done? (Give a restriction on $g$ that ensures that this can be done.)
I think I've managed part 1:
a. We have that $Var(X) = E(X^2) - (E(X))^2$ or alternatively $\sigma^2 = E(X^2) - \mu^2$. The distribution $N(0,1)$ has $\sigma^2 = 1$ and $\mu = 0 \implies \mu^2 = 0$. Hence $E(X^2) = 1$.
b. The Law of Large Numbers states that as $n \to \infty$, the mean of the sample distribution converges to the expectation of the random variable.
I am VERY stuck on number 2... There is a similar question that I've solved
Use the Law of Large Numbers to motivate the approximation to $I = E(g(X))$ with $X \in Un(0, 1)$ and $g(x) = e^{-x^2}$ if $I = \int_0^1 e^{-x^2} dx$
My solution: We write that $I = E(g(X))$ where $X \in Un(0,1)$ and $g(x) = e^{-x^2}$. In terms of integrals, the expectation of $g(X)$ can be written as \begin{align*} E(g(X)) &= \int g(X) \ d F(X) \\ &= \int_0^1 e^{-x^2} \frac{1}{1-0} dx \\ &= \int_0^1 e^{-x^2} dx \\ \end{align*} By the law of large numbers, we know that the mean of a function converges to its expectation. Hence, we can approximate the integral $\int_0^1 e^{-x^2} dx$ by $E(e^{-X^2})$.
Can I use this solution in any way to solve Q2? Am I missing something obvious here? I think the squared random variable is putting me off a bit. I would very much appreciate any hints!
