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Of course! Here are the solutions to the problems. ✅ *** ### 1. Mode of a Gamma Distribution For a random variable $X$ following a Gamma distribution, $X \sim \text{Gamma}(\alpha, \beta)$, with $\alpha > 1$, the probability density function (PDF) is: $$f(x) = \frac{\beta^{\alpha}}{\Gamma(\alpha)} x^{\alpha-1} e^{-\beta x}, \quad x > 0$$ To find the mode, we find the maximum of the PDF by taking the derivative with respect to $x$ and setting it to zero. $$f'(x) = \frac{\beta^{\alpha}}{\Gamma(\alpha)} \left[ (\alpha-1)x^{\alpha-2}e^{-\beta x} - \beta x^{\alpha-1}e^{-\beta x} \right] = 0$$ $$(\alpha-1)x^{\alpha-2}e^{-\beta x} = \beta x^{\alpha-1}e^{-\beta x}$$ Dividing both sides by $x^{\alpha-2}e^{-\beta x}$ (since $x>0$), we get: $$\alpha - 1 = \beta x$$ The mode is **$\mathbf{\frac{\alpha-1}{\beta}}$**. *** ### 2. Mode of a Beta Distribution For a random variable $X \sim \text{Beta}(\alpha, \beta)$, the mode is given by the formula $\frac{\alpha-1}{\alpha+\beta-2}$, provided $\alpha > 1$ and $\beta > 1$. Given $X \sim \text{Beta}(3, 6)$, we have $\alpha=3$ and $\beta=6$. $$\text{mode}(X) = \frac{3-1}{3+6-2} = \frac{2}{7}$$ The mode is **$\mathbf{\frac{2}{7}}$**. *** ### 3. Mode of a Bernoulli Distribution A Bernoulli random variable, $X \sim \text{Ber}(p)$, can take two values: * $P(X=1) = p$ * $P(X=0) = 1-p$ The mode is the value with the highest probability. * If **$p > 0.5$**, then $p > 1-p$, and the mode is **1**. * If **$p < 0.5$**, then $p < 1-p$, and the mode is **0**. * If **$p = 0.5$**, then $p = 1-p$, and both outcomes are equally likely. The distribution is bimodal, and the modes are **0 and 1**. *** ### 4. Mode in a Binary Symmetric Channel The distribution $p_{X|Y}$ represents the posterior probability of the input $X$ given the observed output $Y$. Finding the mode of this distribution for a given $Y=y$ means finding the value of $x$ (0 or 1) that maximizes $P(X=x|Y=y)$. This is known as the **Maximum a Posteriori (MAP)** decision rule. Using Bayes' theorem: $$\hat{x}_{\text{MAP}} = \arg\max_{x \in \{0,1\}} P(X=x|Y=y) = \arg\max_{x \in \{0,1\}} P(Y=y|X=x)P(X=x)$$ Let $P(X=1)=p_1$ and $P(X=0)=1-p_1$. Let the crossover probability be $p$. * If we observe **$Y=0$**, the mode is 1 if $P(X=1|Y=0) > P(X=0|Y=0)$, which simplifies to $p \cdot p_1 > (1-p)(1-p_1)$. Otherwise, the mode is 0. * If we observe **$Y=1$**, the mode is 1 if $P(X=1|Y=1) > P(X=0|Y=1)$, which simplifies to $(1-p)p_1 > p(1-p_1)$. Otherwise, the mode is 0. This is related to the **Maximum Likelihood (ML) heuristic**, which chooses the input $x$ that maximizes the likelihood $P(Y=y|X=x)$, ignoring the prior probabilities $P(X=x)$. The ML rule simplifies to choosing $\hat{x}=y$ (assuming $p0$. We need to find the median $m$ such that $P(X \le m) = 0.5$. The CDF, $F_X(x)$, is: * $F_X(x) = 0$ for $x < 0$. * $F_X(0) = P(X \le 0) = P(X=0) = 0.4$. * For $x > 0$, $F_X(x) = P(X=0) + P(0 < X \le x)$. $$P(0 < X \le x) = \int_0^x 0.6 e^{-t} dt = 0.6[-e^{-t}]_0^x = 0.6(1-e^{-x})$$ So, for $x > 0$, $F_X(x) = 0.4 + 0.6(1-e^{-x})$. Since $F_X(0) = 0.4$, which is less than 0.5, the median must be in the continuous part ($m>0$). We set $F_X(m) = 0.5$: $$0.4 + 0.6(1 - e^{-m}) = 0.5$$ $$0.6(1 - e^{-m}) = 0.1$$ $$1 - e^{-m} = \frac{0.1}{0.6} = \frac{1}{6}$$ $$e^{-m} = 1 - \frac{1}{6} = \frac{5}{6}$$ $$-m = \ln\left(\frac{5}{6}\right) \implies m = -\ln\left(\frac{5}{6}\right) = \ln\left(\frac{6}{5}\right)$$ The median is **$\ln(1.2)$**. rewrite the answers in same way