A random variable $X$ is said to follow a discrete uniform distribution if its probability function is given by
$$p_X(x) = \left\{ \begin{array}{ll}\frac{1}{\theta}, & x = 1, 2, \ldots, \theta\\
0, & \text{otherwise}, \end{array}\right.$$
where $\theta$ is a positive integer. This distribution is denoted $\text{dunif[1, $\theta$]}$ in the following parts of the question.
- Given that $\sum\limits_{j = 1}^n j = \frac{1}{2}n(n + 1)$ and $\sum\limits_{j = 1}^n j^2 = \frac{1}{6} n(n + 1)(2n + 1)$, find the mean and variance of this distribution.
- Determine the cumulative distribution function $F_X(x)$, ensuring that it is defined properly for all $x$.
- The random variables $X$ and $Y$ are independent, with $X \sim \text{dunif}[1, \theta]$ and $Y \sim \text{dunif}[1, \theta]$. Carefully apply the convolution formula to determine the probability function of $W = X + Y$.
- A $20$-sided fair die is rolled as many times as necessary. Write down the probability expression that the third face $10$ turns up on the $n$th roll.
- For a large $n$, what is an approximate distribution for the sample mean $\overline{X}$? Ensure that all the distribution details are given.
I've got the previous parts.
For the last part, my attempt is that it's about CLT (is this correct?) but then I am stuck.
Is the approximate distribution of sample mean Normal here?
And what are the details of such distribution?
Thank you for your help. (This is the complete question)
Best Answer
What you are being asked is as follows. Imagine tossing the die $n$ times. Let $X_i$ be the number obtained on the $i$-th toss, and let $$\bar{X}=\frac{X_1+X_2+\cdots+X_n}{n}$$ For large $n$, what is the approximate distribution of $\bar{X}$?
First we find the mean $\mu$ and the variance $\sigma^2$ of $X_i$.
The random variable $X_i$ has discrete uniform distribution, taking on values $1$ to $20$ each with probability $\frac{1}{20}$. It follows that $$\mu=\frac{1+2+\cdots+20}{20}.$$ By a formula you were given earlier in the problem, the sum $1+2+\cdots +20$ is $\frac{(20)(21)}{2}$. It follows that $\mu=\frac{21}{2}$.
For $\sigma^2$, use the fact this is $E(X_i^2)-(E(X_i))^2$. We need to find $E(X_i^2)$. This is $$\frac{1^2+2^2+\cdots +20^2}{20}.$$ To find the sum of the squares, use the formula given to you for the sum of the first $n$ squares. The sum of the squares is $\frac{(20)(21)(41)}{6}$.
Now you have all the information needed to find $\sigma^2$. Divide the sum of squares by $20$, and subtract $(10.5)^2$.
Finally, the mean of $\bar{X}$ is $\mu$, and the variance of $\bar{X}$ is $\frac{\sigma^2}{n}$. For largish $n$, and it does not have to be terribly large, $\bar{X}$ is approximately normal with mean, variance given in this paragraph. And yes, it is by the Central Limit Theorem that we conclude that $\bar{X}$ has approximately normal distribution.