Suppose that $U_1, U_2, \ldots,U_n $ are independent and identically distributed uniform $(0,1)$ and random variables, and consider their sample mean $$\bar U_n = \frac{1}{n} \sum_{i=1}^n U_i$$

Compute the mean and variance of $\bar{U}$ .

I am confused about what this question is asking. I know the formulas to calculate mean and variance but I don't know how to apply it in this general of a situation.

  • $\begingroup$ The LHS is $\bar{U}$ or $\bar{U}_n$ right? Have you learn the linear property for mean? and variance when they are independent? $\endgroup$ – BGM Oct 13 '17 at 4:50
  • $\begingroup$ U bar n, I was not sure how to format this in my post. And yes I have $\endgroup$ – user469779 Oct 13 '17 at 4:51

Each variable is uniformly distributed, that means

$$ \mathbb{E}[U_i] = \frac{1}{2} $$

Since they are independent

$$ \mathbb{E}[U_n] = \mathbb{E}\left[ \frac{1}{n}\sum_{i=1}^n U_i\right] = \frac{1}{n}\sum_{i=1}^n\mathbb{E}[U_i] = \frac{n}{2n} = \frac{1}{2} $$


$$ \mathbb{V}[U_n] = \mathbb{V}\left[ \frac{1}{n}\sum_{i=1}^n U_i\right] = \frac{1}{n^2}\sum_{i=1}^n\mathbb{V}[U_i] = \cdots $$

Can you take it from here?

  • $\begingroup$ What does the symbol \mathbb{E} mean $\endgroup$ – user469779 Oct 13 '17 at 5:01
  • $\begingroup$ Something weird happened in your expression for the mean. @caverac $\endgroup$ – Michael Oct 13 '17 at 9:06
  • $\begingroup$ @user469779 : $E[X]$ denotes expectation of $X$ ("expectation" is also called "mean"). The key facts here are that for scalars $a_1, ..., a_n$ and any (possibly non-independent) random variables $X_1, ..., X_n$ that have finite means, we have $E[a_1X_1 + ... + a_nX_n] = a_1E[X_1] +...+a_nE[X_n]$. You are trying to compute the mean of $(1/n)U_1 + (1/n)U_2 + ... + (1/n)U_n$. Also, the definition for variance of $X$ is $Var(X)=E[(X-E[X])^2]$, also written as $Var(X)=E[X^2]-E[X]^2$ (whichever form is more convenient). $\endgroup$ – Michael Oct 13 '17 at 9:16
  • $\begingroup$ Would the mean just be the population mean, mu and the variance be sigma/n^2? $\endgroup$ – user469779 Oct 13 '17 at 9:38
  • $\begingroup$ @user469779 Yes, $\mathbb{E}[U_i]=\mu$, the population mean $\endgroup$ – caverac Oct 13 '17 at 10:49

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