et X have the probability distribution \( \mathrm{f}(\mathrm{x})=\frac{1}{4} \times\binom{ 2}{\mathrm{x}} \quad \mathrm{x}=0,1,2 \) Obtain (i) \( M_{x}^{\prime}(t) \) (ii) \( M_{x}^{\prime \prime}(t) \) (iii) \( \operatorname{Var}(\mathrm{x}) \)

To solve for the moment generating function and variance from the given probability distribution \( \mathrm{f}(\mathrm{x})=\frac{1}{4} \times\binom{ 2}{\mathrm{x}} \) for \( \mathrm{x}=0,1,2 \), we start by calculating the moment generating function \( M_x(t) \). ### Step 1: Find \( M_x(t) \) The moment generating function is defined as: \[ M_x(t) = E[e^{tX}] = \sum_{x=0}^{2} e^{tx} f(x) \] Substituting in the values: \[ M_x(t) = \sum_{x=0}^{2} e^{tx} \left( \frac{1}{4} \binom{2}{x} \right) \] Calculating each term: - For \( x=0 \): \( e^{t \cdot 0} \cdot \frac{1}{4} \cdot \binom{2}{0} = 1 \cdot \frac{1}{4} \cdot 1 = \frac{1}{4} \) - For \( x=1 \): \( e^{t \cdot 1} \cdot \frac{1}{4} \cdot \binom{2}{1} = e^t \cdot \frac{1}{4} \cdot 2 = \frac{1}{2} e^t \) - For \( x=2 \): \( e^{t \cdot 2} \cdot \frac{1}{4} \cdot \binom{2}{2} = e^{2t} \cdot \frac{1}{4} \cdot 1 = \frac{1}{4} e^{2t} \) Adding these up gives: \[ M_x(t) = \frac{1}{4} + \frac{1}{2} e^t + \frac{1}{4} e^{2t} \] ### Step 2: Find \( M_x^{\prime}(t) \) Now we differentiate \( M_x(t) \): \[ M_x^{\prime}(t) = \frac{d}{dt}\left( \frac{1}{4} + \frac{1}{2} e^t + \frac{1}{4} e^{2t} \right) \] \[ = 0 + \frac{1}{2} e^t + \frac{1}{4} \cdot 2 e^{2t} = \frac{1}{2} e^t + \frac{1}{2} e^{2t} = \frac{1}{2}(e^t + e^{2t}) \] ### Step 3: Find \( M_x^{\prime \prime}(t) \) Now we differentiate \( M_x^{\prime}(t) \): \[ M_x^{\prime \prime}(t) = \frac{d}{dt}\left( \frac{1}{2}(e^t + e^{2t}) \right) \] \[ = \frac{1}{2}(e^t + 2 e^{2t}) = \frac{1}{2} e^t + e^{2t} \] ### Step 4: Find \( \operatorname{Var}(X) \) The variance can be computed using: \[ \operatorname{Var}(X) = M_x^{\prime \prime}(0) - (M_x^{\prime}(0))^2 \] First, we calculate \( M_x^{\prime}(0) \) and \( M_x^{\prime \prime}(0) \): \[ M_x^{\prime}(0) = \frac{1}{2}(e^0 + e^{0}) = \frac{1}{2}(1 + 1) = 1 \] \[ M_x^{\prime \prime}(0) = \frac{1}{2} e^0 + e^{0} = \frac{1}{2} \cdot 1 + 1 = \frac{3}{2} \] Thus, we find the variance: \[ \operatorname{Var}(X) = M_x^{\prime \prime}(0) - (M_x^{\prime}(0))^2 = \frac{3}{2} - 1^2 = \frac{3}{2} - 1 = \frac{1}{2} \] ### Summary of Results - \( M_x^{\prime}(t) = \frac