Uniform distribution cdf expected value

Question

Given that $X_1, \ldots ,X_n$ are independent random variables, of identical distribution, from a uniform distribution $U(0,10)$, let $\hat{F}(t)$ denotes cdf estimated on a basis of $X_1 \ldots X_n$ in a point t. Find expected value of $\hat{F}(4)$ and variance.

Will $EX$ simply will be $E(0.4)=0.4$? What about the variance?

Answer 1

The empirical distribution function is given by $$ \hat F_n(t)=\frac1n\sum_{i=1}^nI_{\{X_i\le t\}}. $$ Using the linearity of the expectation and the identical distributions, $$ \operatorname E\hat F_n(t)=\frac1n\sum_{i=1}^n\operatorname EI_{\{X_i\le t\}}=\frac1n\sum_{i=1}^nP(X_i\le t)=P(X_1\le t). $$ The variance is given by $$ \operatorname{Var}\hat F_n(t)=\frac1{n^2}\sum_{i=1}^n\operatorname{Var}I_{\{X_i\le t\}}=\frac1nP(X_i\le t)P(X_i>t) $$ since $X_1,\ldots,X_n$ are independent and identically distributed.

Answer 2

By definition, $$\hat{F}(4) = \frac{1}{n}\sum_{i = 1}^n I(X_i \leq 4),$$ where $I$ stands for the indicator function.

By the i.i.d. assumption on $X_1, \ldots, X_n$, $I(X_1 \leq 4), \ldots, I(X_n \leq 4)$ are i.i.d. Bernoulli($p$) random variables. Can you determine $p$ and then proceed?

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For simplicity, denote $I(X_i \leq 4)$ by $Y_i$, $i = 1, \ldots, n$. Then $$p = P(Y_i = 1) = P(X_i \leq 4) = 0.4$$ since $X_i \sim U(0, 10)$. Therefore, $E(Y_i) = p = 0.4$, $\text{Var}(Y_i) = p(1 - p) = 0.4 \times 0.6 = 0.24$, $i = 1, \ldots, n$.

Finally, by independence assumption, $$E(\hat{F}(4)) = \frac{1}{n} \sum_{i = 1}^n E(Y_i).$$ $$\text{Var}(\hat{F}(4)) = \frac{1}{n^2} \sum_{i = 1}^n \text{Var}(Y_i).$$

I think this will be detailed enough.

Answer 3

We have $$\hat{F}(t) = \dfrac{1}{n}\sum_i\mathbb{I}(X_i \leq t)$$ where $\mathbb{I}$ denotes the indicator function.

The expected value of $\hat{F}$ at each $t \in \mathbb{R}$ is $$\mathbb{E}[\hat{F}(t)] = \dfrac{1}{n}\sum_i\mathbb{E}[\mathbb{I}(X_i \leq t)]$$ and it is easy to see that the expected value of an indicator is the probability of the indicator being equal to $1$, i.e., $$\mathbb{E}[\mathbb{I}(X_i \leq t)] = \mathbb{P}(X_i \leq t)\text{.}$$ The variance can be approached similarly, due to independence.