Let $X_1,X_2,...,X_n$ be a sequence of independent and identically distributed random variables from $U(-\sqrt{3},\sqrt{3})$.

Question

Let $X1,X2,...,Xn$ be a sequence of independent and identically distributed random variables from $U(-\sqrt{3},\sqrt{3})$.

Set $n= 100.$

a) What are $E[X_i]$ and $Var(X_i)$ for each $i= 1,...,n$? Answer: $0$

b) Let $Y_n=X_1+···+X_n$. What are $E[Y_n]$ and $Var(Y_n)$ for $n= 100?$ Answer: $0$ and $100$ respectively

c) Using the results above, estimate $P(−16.5< Y_n<16.5)$.

I am stuck on part c. My initial thoughts are to do this: $ \int_{-16.5}^{16.5} f_Y(y_n) \,dy$. However, I'm not exactly sure how to find $f_Y(y_n)$. I know $f_X(x)$ is $\frac{1}{2\sqrt{3}}$. And given that n is $100$, I think $f(y_n)$ is $100\cdot f_X(x)$. Is this the correct approach?

Answer 1

You computed the mean and variance in part (b). So, $$P(-16.5 < Y_n < 16.5) = P\left( \frac{-16.5 - E[Y_n]}{\sqrt{\text{Var}(Y_n)}} < \frac{Y_n - E[Y_n]}{\sqrt{\text{Var}(Y_n)}} < \frac{16.5 - E[Y_n]}{\sqrt{\text{Var}(Y_n)}} \right)$$

When $n$ is large, you can appeal to the central limit theorem to say $Y_n$ is approximately normal. Then $\frac{Y_n - E[Y_n]}{\sqrt{\text{Var}(Y_n)}}$ is approximately standard normal, so you can look up a normal table or use a computer to compute [an approximation of] the probability on the right-hand side above.

Answer 2

The main problem in problem (c) is to estimate $Y_{n}$ probability function. To do so, we use the CTL result which says the following:

\begin{equation*} \text{Let } X_1,X_2,\dots,X_n \text{ be random variables i.i.d. with } E(X_i) = \mu \text{ and } Var(X_i) = \sigma^2 \text{ Then, } \end{equation*} \begin{equation*} \frac{(\sum_{i=1}^{n}X_i)-n\mu}{\sigma\sqrt n} \text{ follows } N(0,1) \hspace{0.2cm} (n\rightarrow +\infty) \end{equation*} The pratical rule used in this scenarios is $n \geq 30$ (which verifies in your case).

So, all you have to do now is to center and reduce $Y_n$ in the given exercise and it's done (by using a table or the calculator).

Answer 3

a) $Var(X_i)=1$. b) $Var(Y_{100})=100$.

The distribution function for $Y_{100}$ is approximately normal with mean and variance given by b) answer. It is NOT $100\times f_X(X)$

Answer 4

Some reality checks are needed here:

First, $E(X_i) = 0, Var(X_i) = 1$, so that $E(Y_{100}) = 0$ and by independence $V(Y_{100}) = 100.$

Second, while $n = 30,$ is a much quoted guideline for assuming that the sum of $n$ random variables is 'nearly' normal by the CLT, this guideline is very often misleading. For uniform random variables $n = 10$ is enough for most purposes and for exponential random variables $n = 100$ is often not enough.

Without working the problem using the CLT, which seems to be the intended method, I will give an approximation by simulation. With 10 million iterations, one can get about three significant digits of accuracy. This is a good match to the approximation using the CLT.

set.seed(521)
y.100 = replicate(10^7, sum(runif(100,-sqrt(3),sqrt(3)))
mean(y.100)
[1] 0.001802827   # aprx 0
var(y.100)
[1] 99.99027      # aprx 100
sd(y.100)
[1] 9.999514      # aprx 10
mean(y.100 > -16.5 & y.100 < 16.5)
[1] 0.9009867     # aprx P(-16.5 < Y.100 < 16.5) (by simulation)

diff(pnorm(c(-16.5,16.5),0,10))
[1] 0.9010571     # aprx P(-16.5 < Y.100 < 16.5) (by CLT aprx)

hdr = "Simulated Dist'n of Sum"
hist(y.100, prob=T, br=30, col="skyblue2", main=hdr)
 curve(dnorm(x, 0, 10), add=T, col="orange", lwd=2)
 abline(v = c(-16.5, 16.5), col="green3", lty = "dotted", lwd=2)