Math Problem Statement

solve

Let Y = X^2, then find the value of E(Y^2) based on the given probability density function.

Solution

We are given a probability density function (PDF) for a random variable XX, and we are asked to find the value of E(Y2)E(Y^2), where Y=X2Y = X^2.

Step 1: Recall the formula for E(Y2)E(Y^2)

Since Y=X2Y = X^2, we can express E(Y2)E(Y^2) as:

E(Y2)=E((X2)2)=E(X4)E(Y^2) = E((X^2)^2) = E(X^4)

Thus, we need to calculate E(X4)E(X^4).

Step 2: Use the formula for the expected value of a function of a random variable

The expected value of a function of a random variable, say g(X)g(X), is given by:

E(g(X))=xg(x)f(x)E(g(X)) = \sum_x g(x) f(x)

In this case, g(X)=X4g(X) = X^4. We will compute E(X4)E(X^4) using the probability distribution provided in the table. Let's list the probabilities:

\hline x & f(x) \\ \hline -2 & \frac{1}{10} \\ -1 & \frac{2}{10} \\ 0 & \frac{1}{10} \\ 1 & \frac{1}{10} \\ 2 & \frac{2}{10} \\ 3 & \frac{1}{10} \\ 4 & \frac{2}{10} \\ \hline \end{array}$$ ### Step 3: Calculate $$E(X^4)$$ We now compute the value of $$X^4$$ for each $$x$$ and multiply it by the corresponding probability: $$E(X^4) = (-2)^4 \cdot \frac{1}{10} + (-1)^4 \cdot \frac{2}{10} + (0)^4 \cdot \frac{1}{10} + (1)^4 \cdot \frac{1}{10} + (2)^4 \cdot \frac{2}{10} + (3)^4 \cdot \frac{1}{10} + (4)^4 \cdot \frac{2}{10}$$ Now calculate: $$E(X^4) = 16 \cdot \frac{1}{10} + 1 \cdot \frac{2}{10} + 0 \cdot \frac{1}{10} + 1 \cdot \frac{1}{10} + 16 \cdot \frac{2}{10} + 81 \cdot \frac{1}{10} + 256 \cdot \frac{2}{10}$$ Simplifying: $$E(X^4) = \frac{16}{10} + \frac{2}{10} + \frac{0}{10} + \frac{1}{10} + \frac{32}{10} + \frac{81}{10} + \frac{512}{10}$$ $$E(X^4) = \frac{16 + 2 + 0 + 1 + 32 + 81 + 512}{10} = \frac{644}{10} = 64.4$$ ### Final Answer: The value of $$E(Y^2) = E(X^4) = 64.4$$. Since none of the options match this exact result, the correct choice is **b. none of these**. --- Would you like more details or have any questions? Here are 5 related questions for further practice: 1. What is the expected value $$E(X)$$ for the given probability distribution? 2. How would you compute the variance $$\text{Var}(X)$$ of this random variable? 3. If $$Y = X^3$$, how would you calculate $$E(Y^2)$$ for this new random variable? 4. How does the shape of the probability distribution affect the value of $$E(X^4)$$? 5. Can you compute $$E(X^2)$$ for this distribution? **Tip:** When dealing with functions of random variables, always check if the problem requires you to transform the variable (like $$X^4$$ in this case) before calculating expectations.

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Math Problem Analysis

Mathematical Concepts

Probability
Expected Value
Random Variables

Formulas

E(g(X)) = Σ g(x) f(x)
E(Y^2) = E(X^4)

Theorems

Expected value of a function of a random variable

Suitable Grade Level

Undergraduate or Advanced High School (Grades 11-12)