Commonly Tested Continuous Distributions (SOA Exam P - Probability - Univariate Random Variables)

[@rmb706]

It took me a bit of head scratching, and it was silly why I was missing the boat. I sure hope that I don't have to do a long sum product on a TI - 30XA though!

[@sadiqueattah1350]

I don't understand how you are calculating the values for PR(Y=0) and so on. Can you please explain?

[@rmb706]

The random variable \( Y \) takes discrete values \(\{0, 1, 2, \dots, 10\}\). To compute the 4th moment of \( Y \), we use the probability mass function of \( Y \): \[ \mathbb{E}[Y^4] = \sum_{y=0}^{10} y^4 \cdot \mathbb{P}(Y = y). \] Since \( Y \) is obtained by rounding \( X \) to the nearest integer, the probabilities for \( Y = y \) are proportional to the length of the intervals for \( X \) that round to \( y \). Specifically: \begin{itemize} \item For \( y = 0 \), \( X \in [0, 0.5) \): \(\mathbb{P}(Y = 0) = \frac{0.5}{10} = 0.05\), \item For \( y = 10 \), \( X \in [9.5, 10] \): \(\mathbb{P}(Y = 10) = \frac{0.5}{10} = 0.05\), \item For \( y = 1, 2, \dots, 9 \), \( X \in [y-0.5, y+0.5) \): \(\mathbb{P}(Y = y) = \frac{1}{10} = 0.1\). \end{itemize} Thus, the probabilities are: \[ \mathbb{P}(Y = y) = \begin{cases} 0.05, & y = 0 \text{ or } y = 10, \\ 0.1, & y = 1, 2, \dots, 9. \end{cases} \] Now compute \( \mathbb{E}[Y^4] \): \[ \mathbb{E}[Y^4] = 0.05 \cdot 0^4 + 0.1 \cdot 1^4 + 0.1 \cdot 2^4 + \cdots + 0.1 \cdot 9^4 + 0.05 \cdot 10^4. \] Expanding the terms: \[ \mathbb{E}[Y^4] = 0.05 \cdot 0 + 0.1 \cdot 1 + 0.1 \cdot 16 + 0.1 \cdot 81 + 0.1 \cdot 256 + 0.1 \cdot 625 + 0.1 \cdot 1296 + 0.1 \cdot 2401 + 0.1 \cdot 4096 + 0.1 \cdot 6561 + 0.05 \cdot 10000. \] Simplify: \[ \mathbb{E}[Y^4] = 0.05 \cdot 10000 + 0.1 \cdot (1 + 16 + 81 + 256 + 625 + 1296 + 2401 + 4096 + 6561). \] Compute the sum inside the parentheses: \[ 1 + 16 + 81 + 256 + 625 + 1296 + 2401 + 4096 + 6561 = 15233. \] Thus: \[ \mathbb{E}[Y^4] = 0.05 \cdot 10000 + 0.1 \cdot 15233 = 500 + 1523.3 = 2023.3. \]