Question

Let $(X_i)_{i\ge 1}$ be a sequence of independent, identically distributed random variables such that \\ $P(X_i = 0) = p$ and $P(X_i = 1) = 1 - p$ \\ for some $p \in [0, 1]$, and let $f: [0, 1] \to \mathbb{R}$ be a continuous function.\\ (i) Prove that \\ $B_n(p) := E\left( f\left(\frac{X_1 + \dots + X_n}{n}\right) \right)$ \\ is a polynomial function of $p$, for any natural number $n$.\\ (ii) Let $\delta > 0$. Prove that \\ $\sum_{k \in K_\delta} \binom{n}{k} p^k (1 - p)^{n - k} \le \frac{1}{4n\delta^2}$, \\ where $K_\delta$ is the set of natural numbers $0 \le k \le n$ such that $|k/n - p| > \delta$.\\ (iii) Show that \\ $\sup_{p \in [0, 1]} |f(p) - B_n(p)| \to 0$ \\ as $n \to \infty$. [You may use without proof that, for any $\epsilon > 0$, there is a $\delta > 0$ such that $|f(x) - f(y)| < \epsilon$ for all $x, y \in [0, 1]$ with $|x - y| < \delta$.]

          Let $(X_i)_{i\ge 1}$ be a sequence of independent, identically distributed random variables such that \\
$P(X_i = 0) = p$ and $P(X_i = 1) = 1 - p$ \\
for some $p \in [0, 1]$, and let $f: [0, 1] \to \mathbb{R}$ be a continuous function.\\
(i) Prove that \\
$B_n(p) := E\left( f\left(\frac{X_1 + \dots + X_n}{n}\right) \right)$ \\
is a polynomial function of $p$, for any natural number $n$.\\ 
(ii) Let $\delta > 0$. Prove that \\
$\sum_{k \in K_\delta} \binom{n}{k} p^k (1 - p)^{n - k} \le \frac{1}{4n\delta^2}$, \\
where $K_\delta$ is the set of natural numbers $0 \le k \le n$ such that $|k/n - p| > \delta$.\\ 
(iii) Show that \\
$\sup_{p \in [0, 1]} |f(p) - B_n(p)| \to 0$ \\
as $n \to \infty$. [You may use without proof that, for any $\epsilon > 0$, there is a $\delta > 0$ such that $|f(x) - f(y)| < \epsilon$ for all $x, y \in [0, 1]$ with $|x - y| < \delta$.]

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Let (Xi)i≥ 1 be a sequence of independent, identically distributed random variables such that

P(Xi = 0) = p and P(Xi = 1) = 1 - p

for some p ∈ [0, 1], and let f: [0, 1] →ℝ be a continuous function.

(i) Prove that

Bn(p) := E( f((X1 + … + Xn)/(n)) )

is a polynomial function of p, for any natural number n.

(ii) Let δ > 0. Prove that

∑k ∈ Kδnk p^k (1 - p)^n - k≤(1)/(4nδ^2),

where Kδ is the set of natural numbers 0 ≤ k ≤ n such that |k/n - p| > δ.

(iii) Show that

supp ∈ [0, 1] |f(p) - Bn(p)| → 0

as n →∞. [You may use without proof that, for any ϵ > 0, there is a δ > 0 such that |f(x) - f(y)| < ϵ for all x, y ∈ [0, 1] with |x - y| < δ.]

Added by Olga A.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Ruirui Liu

09/03/2019

Step 1

.. + X_n}{n}\right)\right) = \sum_{k = 0}^n f\left(\frac{k}{n}\right) P\left(\frac{X_1 + ... + X_n}{n} = \frac{k}{n}\right).$$ Since $X_1, ..., X_n$ are independent and identically distributed, we have $$P\left(\frac{X_1 + ... + X_n}{n} = \frac{k}{n}\right) = Show more…

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Let $(X_i)_{i \ge 1}$ be a sequence of independent, identically distributed random variables such that $$P(X_i = 0) = p \text{ and } P(X_i = 1) = 1 - p$$ for some $p \in [0, 1]$, and let $f: [0, 1] \rightarrow \mathbb{R}$ be a continuous function. (i) Prove that $$B_n(p) := E\left(f\left(\frac{X_1 + ... + X_n}{n}\right)\right)$$ is a polynomial function of $p$, for any natural number $n$. (ii) Let $\delta > 0$. Prove that $$\sum_{k \in K_\delta} \binom{n}{k} p^k (1 - p)^{n - k} \le \frac{1}{4n\delta^2},$$ where $K_\delta$ is the set of natural numbers $0 \le k \le n$ such that $|k/n - p| > \delta$. (iii) Show that $$\sup_{p \in [0, 1]} |f(p) - B_n(p)| \rightarrow 0$$ as $n \rightarrow \infty$. [You may use without proof that, for any $\epsilon > 0$, there is a $\delta > 0$ such that $|f(x) - f(y)| < \epsilon$ for all $x, y \in [0, 1]$ with $|x - y| < \delta$.]

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Transcript

00:06 So, first let's look at the first question a, and we write the variance of this whole term.

00:27 And this theory could be reasonable as the summation of a -i -x -e -i.

00:52 And further ai could be, you know, expanded to a summation of variance of each term.

01:29 And plus the, you know, the coherence of our two terms which are not equal to each other.

01:54 There's i smaller than t.

02:01 A .i.

02:02 A .i.

02:06 Coerance x.

02:25 And since we have the variance of xti.

02:37 So we can have the covariance x t i and x t i so we can have summation a variance also the summation of various a i so we can have this the reason we write this one as this form is that we combine the variance function, or variance of xt, and all those summation of co -awareness error.

04:44 And then now we can separate i and the j summation by written them as i from 1 to n and c.

04:56 C is also from 1 to a, and a, a, g.

05:04 And this one here is auto covariance, tz.

05:19 So we pull along.

05:21 And then look at the question b.

05:31 So from what we've proved in question a, we have this relationship, c from 1 to n.

05:43 So k from 1 to n, n, a, j, a, k, c, s, t, equals the variance of this whole summation.

06:18 And we know that the variance actually is always bigger than zero, so we can't have it.

06:34 This is greater than zero...

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