1. **Problem 1: True or False statements about probability and distributions** (a) Given $B \subset A$ and $P(B) > 0$, check if $P(A|B) \leq P(B|A)$. - Recall conditional probability: $P(A|B) = \frac{P(A \cap B)}{P(B)}$ and $P(B|A) = \frac{P(A \cap B)}{P(A)}$. - Since $B \subset A$, $A \cap B = B$. - So, $P(A|B) = \frac{P(B)}{P(B)} = 1$ and $P(B|A) = \frac{P(B)}{P(A)}$. - Because $P(B) \leq P(A)$, $\frac{P(B)}{P(A)} \leq 1$. - Therefore, $P(A|B) = 1 \geq P(B|A)$, so the statement $P(A|B) \leq P(B|A)$ is FALSE. (b) If $A$ and $B$ are independent, then $A^C$ and $B^C$ are also independent. - Independence means $P(A \cap B) = P(A)P(B)$. - Using complements, $P(A^C \cap B^C) = 1 - P(A) - P(B) + P(A \cap B)$. - Substitute independence: $= 1 - P(A) - P(B) + P(A)P(B) = (1 - P(A))(1 - P(B)) = P(A^C)P(B^C)$. - Hence, $A^C$ and $B^C$ are independent. Statement is TRUE. (c) For independent $X \sim \text{Bin}(n_1, p_1)$ and $Y \sim \text{Bin}(n_2, p_2)$ with $n_1 \neq n_2$, $p_1 \neq p_2$, check if $X + Y \sim \text{Bin}(n_1 + n_2, p_1 + p_2)$. - The sum of independent binomial variables with the same $p$ is binomial with summed $n$. - Here $p_1 \neq p_2$, so $X+Y$ is NOT binomial with parameter $p_1 + p_2$. - Statement is FALSE. (d) For independent $X \sim N(\mu_1, \sigma_1^2)$ and $Y \sim N(\mu_2, \sigma_2^2)$, check if $X + Y \sim N(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2)$. - Sum of independent normal variables is normal with mean sum and variance sum. - Statement is TRUE. 2. **Problem 2: True or False statements about sample mean and variance distributions** Given $X_i \sim N(\mu, \sigma^2)$ i.i.d., $\bar{X} = \frac{1}{n} \sum X_i$, and $S^2 = \frac{1}{n-1} \sum (X_i - \bar{X})^2$. (a) $\bar{X} \sim N(\mu, \frac{\sigma^2}{n})$. - The sample mean of normal variables is normal with mean $\mu$ and variance $\sigma^2/n$. - TRUE. (b) $\frac{(n-1)S^2}{\sigma^2} \sim \chi_n^2$. - The scaled sample variance follows $\chi_{n-1}^2$ distribution, not $\chi_n^2$. - FALSE. (c) $\sum \frac{(X_i - \mu)^2}{\sigma^2} \sim \chi_n^2$. - Sum of squared standardized normal variables is $\chi_n^2$. - TRUE. (d) $\frac{\sqrt{n}(\bar{X} - \mu)}{S} \sim t_{n-1}$. - This is the definition of the t-statistic with $n-1$ degrees of freedom. - TRUE. 3. **Problem 3: Expectation calculations for joint pdf** Given joint pdf: $$f(x_1, x_2) = \frac{2}{\theta^2} e^{-\frac{x_1 + x_2}{\theta}}, \quad 0 < x_1 < x_2 < \infty, \theta > 0$$ (a) Find $E(X_1)$. - First find marginal pdf of $X_1$: $$f_{X_1}(x_1) = \int_{x_2 = x_1}^\infty f(x_1, x_2) dx_2 = \int_{x_2 = x_1}^\infty \frac{2}{\theta^2} e^{-\frac{x_1 + x_2}{\theta}} dx_2 = \frac{2}{\theta^2} e^{-\frac{x_1}{\theta}} \int_{x_2 = x_1}^\infty e^{-\frac{x_2}{\theta}} dx_2$$ - Evaluate integral: $$\int_{x_1}^\infty e^{-\frac{x_2}{\theta}} dx_2 = \theta e^{-\frac{x_1}{\theta}}$$ - So, $$f_{X_1}(x_1) = \frac{2}{\theta^2} e^{-\frac{x_1}{\theta}} \theta e^{-\frac{x_1}{\theta}} = \frac{2}{\theta} e^{-\frac{2x_1}{\theta}}, \quad x_1 > 0$$ - This is an exponential distribution with rate $\lambda = \frac{2}{\theta}$. - Hence, $$E(X_1) = \frac{1}{\lambda} = \frac{\theta}{2}$$ (b) Find $E(X_2 | X_1 = x_1)$. - Conditional pdf: $$f_{X_2|X_1}(x_2|x_1) = \frac{f(x_1, x_2)}{f_{X_1}(x_1)} = \frac{\frac{2}{\theta^2} e^{-\frac{x_1 + x_2}{\theta}}}{\frac{2}{\theta} e^{-\frac{2x_1}{\theta}}} = \frac{1}{\theta} e^{-\frac{x_2 - x_1}{\theta}}, \quad x_2 > x_1$$ - This is an exponential distribution shifted by $x_1$ with rate $1/\theta$. - So, $$E(X_2 | X_1 = x_1) = x_1 + \theta$$ (c) Find $E(X_1 X_2)$. - Use law of total expectation: $$E(X_1 X_2) = E\left[X_1 E(X_2 | X_1)\right] = E[X_1 (X_1 + \theta)] = E(X_1^2) + \theta E(X_1)$$ - For exponential with rate $\lambda = \frac{2}{\theta}$: $$E(X_1) = \frac{1}{\lambda} = \frac{\theta}{2}$$ $$Var(X_1) = \frac{1}{\lambda^2} = \frac{\theta^2}{4}$$ - So, $$E(X_1^2) = Var(X_1) + (E(X_1))^2 = \frac{\theta^2}{4} + \left(\frac{\theta}{2}\right)^2 = \frac{\theta^2}{4} + \frac{\theta^2}{4} = \frac{\theta^2}{2}$$ - Therefore, $$E(X_1 X_2) = \frac{\theta^2}{2} + \theta \cdot \frac{\theta}{2} = \frac{\theta^2}{2} + \frac{\theta^2}{2} = \theta^2$$ **Final answers:** - Q1: (a) FALSE, (b) TRUE, (c) FALSE, (d) TRUE - Q2: (a) TRUE, (b) FALSE, (c) TRUE, (d) TRUE - Q3: (a) $\frac{\theta}{2}$, (b) $x_1 + \theta$, (c) $\theta^2$