This is a problem involving the Weibull distribution and calculating the expected value of a transformed random variable. Here's the breakdown:

Problem Statement:

The random variable $X$ follows the Weibull distribution with parameters $\alpha = 2$ and $\lambda = 5$ (a special case known as the Rayleigh distribution). Another random variable $Y$ is defined as $Y = X^2$. The goal is to find the mathematical expectation $\mathbb{E}[Y]$, which is the expected value of $Y$.

Step-by-Step Solution:

1. Probability Density Function (PDF) of $X$: For a Weibull distribution with shape $\alpha$ and scale $\lambda$, the PDF is: $f_X(x) = \frac{\alpha}{\lambda^\alpha} x^{\alpha - 1} e^{-(x/\lambda)^\alpha}, \quad x \geq 0$ Substituting $\alpha = 2$ and $\lambda = 5$: $f_X(x) = \frac{2}{5^2} x^{2-1} e^{-(x/5)^2} = \frac{2}{25} x e^{-(x^2/25)}, \quad x \geq 0$

2. Relation Between $X$ and $Y$: Since $Y = X^2$, the transformation is: $Y = X^2 \implies X = \sqrt{Y}, \quad x \geq 0 \implies y \geq 0$

3. Expected Value of $Y$: By definition: $\mathbb{E}[Y] = \int_0^\infty y f_Y(y) \, dy$ However, it’s easier to compute $\mathbb{E}[Y]$ using the transformation and the known PDF of $X$: [ \mathbb{E}[Y] = \mathbb{E}[X^2] = \int_0^\infty x^2 f_X(x) , dx ]

4. Substitute $f_X(x)$: Using the PDF $f_X(x) = \frac{2}{25} x e^{-(x^2/25)}$: $\mathbb{E}[Y] = \int_0^\infty x^2 \cdot \frac{2}{25} x e^{-(x^2/25)} \, dx = \frac{2}{25} \int_0^\infty x^3 e^{-(x^2/25)} \, dx$

5. Simplify the Integral: Let $u = \frac{x^2}{25}$, so $x^2 = 25u$ and $dx = \frac{25}{2\sqrt{25u}} \, du = \frac{5}{\sqrt{u}} \, du$: $\mathbb{E}[Y] = \frac{2}{25} \int_0^\infty (25u) \cdot 5 \cdot e^{-u} \, du = \frac{2}{25} \cdot 125 \int_0^\infty u e^{-u} \, du$

6. Evaluate the Gamma Function: The integral $\int_0^\infty u e^{-u} \, du$ is a standard gamma function $\Gamma(2)$, where $\Gamma(n) = (n-1)!$ for integer $n$. Here: $\Gamma(2) = 1! = 1$ Thus: $\mathbb{E}[Y] = \frac{2}{25} \cdot 125 \cdot 1 = 10$

Final Answer:

The expected value of $Y$ is: $\mathbb{E}[Y] = 10$

Would you like a deeper explanation of any step or help with related problems? Here are five follow-up questions to expand your understanding:

1. What is the general formula for transforming expectations when $Y = g(X)$?
2. How does the Weibull distribution generalize to other distributions (e.g., Rayleigh)?
3. What is the role of the gamma function in expectation calculations for Weibull-distributed variables?
4. How can this result be verified using numerical integration or simulation?
5. What are the applications of the Weibull and Rayleigh distributions in real-world contexts?

Tip: When working with transformed random variables, always check if the transformation simplifies the problem (e.g., directly relating moments like $\mathbb{E}[X^2]$).