/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 Let \(f\left(x_{1}, x_{2}\right)... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 3

Let \(f\left(x_{1}, x_{2}\right)=21 x_{1}^{2} x_{2}^{3}, 0

Short Answer

Expert verified
The conditional mean of \(X_1\) given \(X_2=x_2\) is \(\frac{1}{2}x_{2}\), the conditional variance is \(\frac{1}{12}x_2^{2}\). The pdf of \(Y=E\left(X_{1} \mid X_{2}\right)\) is \(f_Y(y) = 28y^2\). The expectation of \(Y\) is \(\frac{7}{10}\) and its variance is \(\frac{7}{150}\). Both \(E(Y)\) and \(\operatorname{Var}(Y)\) are smaller than \(E(X_{1})\) and \(\operatorname{Var}(X_{1})\) respectively.

Step by step solution

01

Calculation of Conditional Probability Density Function

Firstly, find the marginal pdf of \(X_2\) by integrating the joint pdf over \(x_1\) from 0 to \(x_2\). We get \(f_2\left(x_{2}\right)= 14x_{2}^{2}\). The conditional pdf of \(X_1\) given \(X_2 = x_2\) is then given by \(f_{X_1|X_{2}=x_{2}}(x_1) = \frac{f\left(x_{1}, x_{2}\right)}{f_{2}\left(x_{2}\right)} = \frac{21 x_{1}^{2} x_{2}^{3}}{14x_{2}^{2}} = 3x_{1}^{2}x_{2}\), for \(0<x_{1}<x_{2}\).
02

Calculation of Conditional Mean and Variance of \(X_1\) Given \(X_2=x_2\)

The conditional mean is given by \(\mu_{X_{1}|X_{2}=x_{2}} = \int_0^{x_2} x_{1}f_{X_1|X_{2}=x_{2}}(x_1) dx_{1} = \frac{1}{2}x_{2}\). The conditional variance is given by \(\sigma^{2}_{X_{1}|x_{2}} = \int_0^{x_2} (x_1-\mu_{X_1|x_2})^{2}f_{X_1|X_{2}=x_{2}}(x_1) dx_{1} = \frac{1}{12}x_2^{2}\).
03

Finding the Distribution of \(Y=E(X_{1}|X_{2})\)

Here, \(Y=E(X_{1}|X_{2}) = \frac{1}{2}X_{2}\) is the conditional mean derived in the previous step. Being a function of \(X_2\), its distribution is the same as the marginal distribution of \(X_{2}\), and its pdf is therefore \(f_Y(y) = 28y^2\), for \(0<y<1\).
04

Determine \(E(Y)\) and \(\operatorname{Var}(Y)\)

We can compute the expectation of \(Y\) as \(\mathbb{E}(Y) = \int_0^1 yf_Y(y) dy = \frac{7}{10}\). The variance of \(Y\) can be computed as \(\operatorname{Var}(Y) = \int_0^1 (y-\mathbb{E}(Y))^2f_Y(y) dy = \frac{7}{150}\).
05

Comparing with \(E(X_{1})\) and \(\operatorname{Var}(X_{1})\)

We compute \(\mathbb{E}(X_{1})\), the expectation of \(X_{1}\), as the double integral \(\int_0^1 \int_0^{x_2} x_{1} f(x_1,x_2)dx_1 dx_2 = \frac{7}{15}\) which is larger than \(\mathbb{E}(Y)\). Similar computation for \(\operatorname{Var}(X_{1})\) gives \(\int_0^1 \int_0^{x_2} (x_1-\mathbb{E}(X_{1}))^{2} f(x_1,x_2) dx_1 dx_2 = \frac{7}{90}\) which is also larger than \(\operatorname{Var}(Y)\). Thus, \(E(Y) < E(X_{1})\) and \(\operatorname{Var}(Y) < \operatorname{Var}(X_{1})\).

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