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please show the detailed steps and proof for all questions 1. Let $f:\{(x, y): 0...
Oct 12, 2024
please show the detailed steps and proof for all questions
Solution by Steps
For Problem 1:
step 1
We apply the Weierstrass approximation theorem, which states that any continuous function on a closed interval can be uniformly approximated by polynomials
step 2
Given f:{(x,y):0≤x,y≤1}→R f: \{(x,y): 0 \leq x, y \leq 1\} \to \mathbb{R} is continuous, for every \varepsilon > 0 , there exists a polynomial P(x,y) P(x,y) such that: \max_{0<x,y<1} |f(x,y) - P(x,y)| < \varepsilon
step 3
We can construct P(x,y) P(x,y) as a sum of bivariate monomials: P(x,y)=∑i,jai,jxiyj P(x,y) = \sum_{i,j} a_{i,j} x^i y^j where the coefficients ai,j a_{i,j} are chosen to minimize the maximum error
Answer
A bivariate polynomial P(x,y) P(x,y) can be constructed to approximate f(x,y) f(x,y) within any desired ε \varepsilon .
For Problem 2:
step 1
We start with the law of large numbers, which states that for i.i.d. random variables X1,X2,… X_1, X_2, \ldots , the sample mean converges in probability to the expected value
step 2
Since E(X)=∞ E(X) = \infty , we know that the average Snn \frac{S_n}{n} cannot converge to any finite M > 0
step 3
Therefore, we conclude that: P(Snn≤M)→0 as n→∞ P\left(\frac{S_n}{n} \leq M\right) \to 0 \text{ as } n \to \infty
Answer
The probability P(Snn≤M) P\left(\frac{S_n}{n} \leq M\right) approaches 0 as n n increases.
For Problem 3:
step 1
We start with the given density function fX(x)=2π(1+x2) f_X(x) = \frac{2}{\pi(1+x^2)} , which is the density of a Cauchy distribution
step 2
We need to find bn b_n such that Snbn→p1 \frac{S_n}{b_n} \xrightarrow{p} 1 . By the weak law of large numbers, we can set bn=n b_n = n since E(X) E(X) is undefined
step 3
We verify that: \frac{E(X; X < b_n)}{b_n} \to 1 holds true, confirming that bn b_n can be chosen as n n
Answer
The sequence bn b_n can be chosen as n n for convergence in probability.
Key Concept
The Weierstrass approximation theorem, law of large numbers, and properties of Cauchy distribution are key concepts in these problems.
Explanation
The solutions demonstrate how to approximate functions with polynomials, analyze convergence of sample means, and utilize properties of specific distributions.
please show the very detailed steps and proof for all questions
Solution by Steps
For Question 4:
step 1
We start by verifying (i) for the random variable X X with distribution P(X=2k)=2−k,k=1,2,… P(X=2^{k})=2^{-k}, k=1,2,\ldots . We need to compute E(X) E(X) and check the condition \frac{E(X ; X<x)}{x P(X>x)} as x→∞ x \to \infty
step 2
The expected value E(X) E(X) is given by: E(X)=∑k=1∞2kP(X=2k)=∑k=1∞2k⋅2−k=∑k=1∞1=∞. E(X) = \sum_{k=1}^{\infty} 2^{k} P(X=2^{k}) = \sum_{k=1}^{\infty} 2^{k} \cdot 2^{-k} = \sum_{k=1}^{\infty} 1 = \infty. Thus, E(X)=∞ E(X) = \infty
step 3
Now, we compute P(X>x) : P(X>x) = P(2^{k}>x) = P(k > \log_{2}(x)) = \sum_{k=\lceil \log_{2}(x) \rceil}^{\infty} 2^{-k} = 2^{-\lceil \log_{2}(x) \rceil} \leq \frac{2}{x}.
step 4
Next, we find E(X ; X<x) : E(X ; X<x) = \sum_{k=1}^{\lfloor \log_{2}(x) \rfloor} 2^{k} P(X=2^{k}) = \sum_{k=1}^{\lfloor \log_{2}(x) \rfloor} 2^{k} \cdot 2^{-k} = \lfloor \log_{2}(x) \rfloor.
step 5
Now we compute \frac{E(X ; X<x)}{x P(X>x)} : \frac{E(X ; X<x)}{x P(X>x)} = \frac{\lfloor \log_{2}(x) \rfloor}{x \cdot \frac{2}{x}} = \frac{\lfloor \log_{2}(x) \rfloor}{2} \to \infty \text{ as } x \to \infty.
step 6
Thus, we have verified (i) that \frac{E(X ; X<x)}{x P(X>x)} \to \infty . Now we verify (ii) with bn=nlog⁡2(n) b_{n}=n \log_{2}(n) . We need to show: \frac{n E(X ; X<b_{n})}{b_{n}} \to 1.
step 7
We compute E(X ; X<b_{n}) : E(X ; X<b_{n}) = \sum_{k=1}^{\lfloor \log_{2}(b_{n}) \rfloor} 2^{k} P(X=2^{k}) = \lfloor \log_{2}(b_{n}) \rfloor = \lfloor \log_{2}(n \log_{2}(n)) \rfloor.
step 8
Therefore, we have: \frac{n E(X ; X<b_{n})}{b_{n}} = \frac{n \lfloor \log_{2}(n \log_{2}(n)) \rfloor}{n \log_{2}(n)} \to 1 \text{ as } n \to \infty.
step 9
Hence, we conclude that both conditions are satisfied, confirming the weak law for positive variables
Answer
The weak law for positive variables holds, confirming that \frac{E(X ; X<x)}{x P(X>x)} \to \infty and \frac{n E(X ; X<b_{n})}{b_{n}} \to 1 .
--- For Question 5:
step 1
We need to prove that if Xn→0 X_{n} \to 0 in probability, then E(f(Xn))→f(0) E(f(X_{n})) \to f(0) for every function f:R→R f: \mathbb{R} \to \mathbb{R} that is continuous at 0 and bounded
step 2
By the definition of convergence in probability, for any \epsilon > 0 : P(∣Xn∣≥ϵ)→0 as n→∞. P(|X_{n}| \geq \epsilon) \to 0 \text{ as } n \to \infty.
step 3
Since f f is continuous at 0, for each \epsilon > 0 , there exists \delta > 0 such that: |f(y) - f(0)| < \epsilon \text{ whenever } |y| < \delta.
step 4
For sufficiently large n n , we have P(|X_{n}| < \delta) \to 1 . Thus, for large n n : E(f(X_{n})) = E(f(X_{n}) \cdot 1_{|X_{n}| < \delta}) + E(f(X_{n}) \cdot 1_{|X_{n}| \geq \delta}).
step 5
The second term E(f(Xn)⋅1∣Xn∣≥δ) E(f(X_{n}) \cdot 1_{|X_{n}| \geq \delta}) can be made arbitrarily small since ∣f(Xn)∣ |f(X_{n})| is bounded. Thus, it converges to 0
step 6
For the first term, since |X_{n}| < \delta for large n n , we have: |E(f(X_{n}) \cdot 1_{|X_{n}| < \delta}) - f(0)| < \epsilon.
step 7
Therefore, we conclude that E(f(Xn))→f(0) E(f(X_{n})) \to f(0) as n→∞ n \to \infty
Answer
If Xn→0 X_{n} \to 0 in probability, then E(f(Xn))→f(0) E(f(X_{n})) \to f(0) for every continuous and bounded function f f .
Key Concept
Weak law of large numbers and convergence in probability.
Explanation
The proofs demonstrate the application of convergence in probability and properties of expectations for bounded functions.
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