Absolutely! Let's walk through each part of **Problem 4: Median, Mode** step by step, explaining the concepts and how to approach each one. --- ### 1. **Mode of Gamma Distribution \((\alpha, \beta)\), \(\alpha > 1\):** **Gamma distribution:** \(f(x) = \frac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{-\beta x}\), for \(x > 0\). - **Mode:** The value of \(x\) where the pdf is maximized. - For \(\alpha > 1\), the mode is at: \[ \text{mode}(X) = \frac{\alpha-1}{\beta} \] --- ### 2. **Mode of Beta Distribution \(\text{Beta}(3,6)\):** **Beta distribution:** \(f(x) = \frac{x^{a-1}(1-x)^{b-1}}{B(a,b)}\), \(x \in (0,1)\). - **Mode:** For \(a > 1, b > 1\), the mode is: \[ \text{mode}(X) = \frac{a-1}{a+b-2} \] - Here, \(a=3, b=6\): \[ \text{mode}(X) = \frac{3-1}{3+6-2} = \frac{2}{7} \] --- ### 3. **Mode for Bernoulli Distribution \(\text{Ber}(p)\):** - \(X = 1\) with probability \(p\), \(X = 0\) with probability \(1-p\). - **Mode:** The value (0 or 1) with the higher probability: - If \(p > 0.5\), mode is 1. - If \(p < 0.5\), mode is 0. - If \(p = 0.5\), both 0 and 1 are modes (bimodal). --- ### 4. **Mode of \(p_{X|Y}\) in Binary Symmetric Channel:** - In a binary symmetric channel, the posterior \(p_{X|Y}\) is maximized by the most likely \(X\) given \(Y\). - **Mode:** The value of \(X\) that maximizes \(p_{X|Y}\). - **Relation to heuristic:** This is the MAP (Maximum A Posteriori) estimate: choose the value of \(X\) with the highest posterior probability given \(Y\). --- ### 5. **Median of Exponential Distribution \(\text{Exp}(\lambda)\):** **Exponential distribution:** \(f(x) = \lambda e^{-\lambda x}, x\geq 0\). - **Median:** Solve \(P(X \leq m) = 0.5\): \[ 1 - e^{-\lambda m} = 0.5 \implies e^{-\lambda m} = 0.5 \implies -\lambda m = \ln(0.5) \] \[ m = \frac{\ln 2}{\lambda} \] --- ### 6. **Random Variable with Finite Mean but Infinite Variance:** - **Example:** Pareto distribution with parameter \(\alpha \in (1,2)\). - **Mean:** \(\frac{\alpha x_m}{\alpha-1}\), finite for \(\alpha > 1\). - **Variance:** Infinite for \(\alpha \leq 2\). - **Median:** \(x_m 2^{1/\alpha}\), where \(x_m\) is the scale parameter. --- ### 7. **Median of Mixed Distribution:** Given: - \(P(X=0)=0.4\) - \(f_X(x)=0.6 e^{-x}\), \(x>0\) (so \(P(X>0)=0.6\)) **Find median \(m\):** - Median is the value where \(P(X \leq m) = 0.5\). - For \(m = 0\): \(P(X \leq 0) = 0.4 < 0.5\), so median \(>0\). - For \(m > 0\): \[ P(X \leq m) = 0.4 + 0.6(1 - e^{-m}) = 0.4 + 0.6 - 0.6e^{-m} = 1 - 0.6e^{-m} \] Set equal to 0.5: \[ 1 - 0.6e^{-m} = 0.5 \implies 0.6e^{-m} = 0.5 \implies e^{-m} = \frac{5}{6} \] \[ m = -\ln\left(\frac{5}{6}\right) \] --- If you need detailed calculations or further explanation on any specific part, let me know!