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 and variance of \(X_{1}\) given \(X_{2}=x_{2}\) are \(\frac{x_{2}}{3}\) and \(\frac{x_{2}^2}{18}\) respectively. The distribution of \(Y=E[X_{1}|X_{2}]\) is \(\frac{X_{2}}{3}\). The expectation and variance of \(Y\) are \(7/18\) and \(1/54\). Comparing it with \(X_{1}\), we find that \(E[X_{1}]\) is smaller than \(E(Y)\) while \(\operatorname{var}(X_{1})\) is larger than \(\operatorname{var}(Y)\).
Step by step solution
01
Calculate marginal pdf of \(X_{2}\)
Firstly, the marginal probability density function (pdf) of \(X_{2}\) needs to be calculated. This is done by integrating the joint pdf \(f(x_{1}, x_{2})\) over all possible values of \(x_{1}\). The result is \(f_{X_{2}}(x_{2}) = 7x_{2}^{2}\) for \(0<x_{2}<1\), zero elsewhere.
02
Calculate Conditional pdf of \(X_{1}\) given \(X_{2}=x_{2}\)
Next, the conditional pdf of \(X_{1}\) given \(X_{2}=x_{2}\) is calculated using the formula of conditional probability which is the ratio of joint pdf to the marginal pdf of \(X_{2}\). The result is \(f_{X_{1}|X_{2}}(x_{1}|x_{2}) = 3x_{1}^{2}\) for \(0<x_{1}<x_{2}\), zero elsewhere.
03
Find the Conditional Mean and Variance of \(X_{1}\) given \(X_{2}=x_{2}\)
Now we can calculate the conditional mean and variance of \(X_{1}\) given \(X_{2}=x_{2}\). The conditional mean is the expected value of \(X_{1}\) given \(X_{2}=x_{2}\), which is \(E[X_{1}|X_{2}=x_{2}]=\int_0^{x_{2}}x_{1}f_{X_{1}|X_{2}}(x_{1}|x_{2}) dx_1 = \frac{x_{2}}{3}\). The conditional variance is the expected value of the square of \(X_{1}\) minus the square of the expected value of \(X_{1}\) and it is obtained as \(Var[X_{1}|X_{2}=x_{2}]=\int_0^{x_{2}}x_{1}^2f_{X_{1}|X_{2}}(x_{1}|x_{2}) dx_1 - \left(E[X_{1}|X_{2}=x_{2}]\right)^2 = \frac{x_{2}^2}{18}\).
04
Find the Distribution of \(Y=E(X_{1} \mid X_{2})\)
The distribution of \(Y=E[X_{1}|X_{2}]\) can be obtained by substituting the result from step 3 so we get \(Y = \frac{X_{2}}{3}\). This means Y follows the distribution of \(X_{2}\) divided by 3.
05
Determine \(E(Y)\) and \(\operatorname{var}(Y)\)
The expectation \(E(Y)\) and variance of \(Y\) can be calculated using the derived distribution from step 4. We find that \(E(Y) = E(X_{2})/3 = 7/18\) and \(\operatorname{var}(Y) = \operatorname{var}(X_{2})/9 = 1/54\).
06
Compare These to \(E(X_{1})\) and \(\operatorname{var}(X_{1})\)
Finally, we have to calculate \(E(X_{1})\) and \(\operatorname{var}(X_{1})\) and compare these with the results from step 5. We find that \(E(X_{1})= \int_0^{1}\int_0^{x_{2}} x_{1}f(x_{1}, x_{2}) dx_{1} dx_{2} = 1/8\), \(\operatorname{var}(X_{1}) = \int_0^{1}\int_0^{x_{2}} x_{1}^2f(x_{1}, x_{2}) dx_{1} dx_{2} - \left(E[X_{1}]\right)^2 = 7/240\). Comparing with the results from step 5, we find that \(E[X_{1}]\) is smaller than \(E(Y)\) while \(\operatorname{var}(X_{1})\) is larger than \(\operatorname{var}(Y)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Marginal Distribution
A marginal distribution provides insight into the distribution of one random variable by itself from a joint distribution of two or more variables. In our exercise, we have a joint probability density function (pdf) for two random variables, denoted as \(X_1\) and \(X_2\). To find the marginal distribution of one of these variables, say \(X_2\), we integrate the joint pdf over the values of the other variable, \(X_1\).
This process essentially "marginalizes" \(X_1\) out of the joint distribution, leaving us with a probability density function that describes \(X_2\) alone. By solving \[\int_0^{x_2} 21 x_1^2 x_2^3 \, dx_1\]we find the marginal pdf of \(X_2\) to be \(7x_2^2\) for \( 0 < x_2 < 1 \), which tells us how likely different values of \(X_2\) are, without considering \(X_1\). The marginal distribution is instrumental in deriving other statistical measures, such as conditional probabilities.
This process essentially "marginalizes" \(X_1\) out of the joint distribution, leaving us with a probability density function that describes \(X_2\) alone. By solving \[\int_0^{x_2} 21 x_1^2 x_2^3 \, dx_1\]we find the marginal pdf of \(X_2\) to be \(7x_2^2\) for \( 0 < x_2 < 1 \), which tells us how likely different values of \(X_2\) are, without considering \(X_1\). The marginal distribution is instrumental in deriving other statistical measures, such as conditional probabilities.
Joint Probability Density Function
The joint probability density function (pdf) is a key concept when we study continuous random variables. It represents the probability that the variables take on a particular set of values. In the context of our exercise, \(f(x_1, x_2) = 21x_1^2x_2^3\) describes how \(X_1\) and \(X_2\) are distributed together over the given range \(0 < x_1 < x_2 < 1\).
A joint pdf must satisfy two conditions:
A joint pdf must satisfy two conditions:
- It must be non-negative for all values within the defined range.
- The integral over the entire space must equal 1, ensuring it represents a valid probability distribution.
Conditional Variance
Conditional variance is a measure of how much variability there is in one random variable, given the value of another random variable. In the given exercise, we focus on the conditional variance of \(X_1\) given \(X_2 = x_2\).
The formula to find the conditional variance is:\[\operatorname{Var}(X_1 | X_2 = x_2) = E[X_1^2|X_2 = x_2] - (E[X_1|X_2 = x_2])^2\]To find \(E[X_1^2|X_2 = x_2]\), we calculate:\[\int_0^{x_2} x_1^2 \cdot 3x_1^2 \, dx_1\]Using these computations, we obtain the conditional variance: \(\frac{x_2^2}{18}\).
Conditional variance is important because it tells us how much uncertainty remains in \(X_1\) after accounting for the information provided by \(X_2 = x_2\). Understanding conditional relationships helps in better predictions and understanding the mechanisms behind certain processes.
The formula to find the conditional variance is:\[\operatorname{Var}(X_1 | X_2 = x_2) = E[X_1^2|X_2 = x_2] - (E[X_1|X_2 = x_2])^2\]To find \(E[X_1^2|X_2 = x_2]\), we calculate:\[\int_0^{x_2} x_1^2 \cdot 3x_1^2 \, dx_1\]Using these computations, we obtain the conditional variance: \(\frac{x_2^2}{18}\).
Conditional variance is important because it tells us how much uncertainty remains in \(X_1\) after accounting for the information provided by \(X_2 = x_2\). Understanding conditional relationships helps in better predictions and understanding the mechanisms behind certain processes.