1. **Problem Statement:** A random variable $X$ has a probability distribution given by $$f(x) = p(1-p)^x \text{ for } x=0,1,2,\ldots \text{ and } 0 \text{ otherwise}.$$ (i) Derive the moment generating function (MGF) of $X$. 2. **Formula and Explanation:** The moment generating function $M_X(t)$ is defined as $$M_X(t) = E[e^{tX}] = \sum_{x=0}^\infty e^{tx} f(x).$$ Since $f(x) = p(1-p)^x$, substitute to get $$M_X(t) = \sum_{x=0}^\infty e^{tx} p (1-p)^x = p \sum_{x=0}^\infty \left(e^t (1-p)\right)^x.$$ This is a geometric series with ratio $r = e^t (1-p)$, which converges if $|r| < 1$. 3. **Intermediate Work:** Using the geometric series sum formula $$\sum_{x=0}^\infty r^x = \frac{1}{1-r},$$ we get $$M_X(t) = \frac{p}{1 - e^t (1-p)} \quad \text{for } |e^t (1-p)| < 1.$$ 4. **Mean and Variance from MGF:** The mean $E[X]$ and variance $Var(X)$ can be found from derivatives of $M_X(t)$: $$E[X] = M_X'(0), \quad Var(X) = M_X''(0) - (M_X'(0))^2.$$ Calculate the first derivative: $$M_X(t) = \frac{p}{1 - (1-p)e^t} = p \left(1 - (1-p)e^t\right)^{-1}.$$ Using the chain rule, $$M_X'(t) = p \cdot (-1) \cdot \left(1 - (1-p)e^t\right)^{-2} \cdot (-(1-p)e^t) = \frac{p (1-p) e^t}{\left(1 - (1-p)e^t\right)^2}.$$ Evaluate at $t=0$: $$M_X'(0) = \frac{p (1-p)}{(1 - (1-p))^2} = \frac{p (1-p)}{p^2} = \frac{1-p}{p}.$$ Calculate the second derivative: $$M_X''(t) = \frac{d}{dt} M_X'(t) = \frac{d}{dt} \left( \frac{p (1-p) e^t}{(1 - (1-p)e^t)^2} \right).$$ Using quotient and chain rules, $$M_X''(t) = \frac{p (1-p) e^t (1 - (1-p)e^t)^2 + p (1-p) e^t \cdot 2 (1 - (1-p)e^t) (1-p) e^t}{(1 - (1-p)e^t)^4}$$ Simplify numerator at $t=0$: $$M_X''(0) = \frac{p (1-p) (1)^2 + 2 p (1-p)^2}{1^4} = p (1-p) + 2 p (1-p)^2 = p (1-p)(1 + 2(1-p)).$$ Calculate explicitly: $$1 + 2(1-p) = 1 + 2 - 2p = 3 - 2p,$$ so $$M_X''(0) = p (1-p)(3 - 2p).$$ Divide by $p^3$ due to denominator powers (rechecking carefully): Actually, the denominator at $t=0$ is $(1 - (1-p) e^0)^4 = (1 - (1-p) imes 1)^4 = p^4$. Recalculate numerator at $t=0$: First term: $p (1-p) e^0 (1 - (1-p) e^0)^2 = p (1-p) imes 1 imes p^2 = p^3 (1-p)$ Second term: $2 p (1-p)^2 e^{2 imes 0} (1 - (1-p) e^0) = 2 p (1-p)^2 imes 1 imes p = 2 p^2 (1-p)^2$ Sum numerator: $p^3 (1-p) + 2 p^2 (1-p)^2 = p^2 (1-p) (p + 2 (1-p)) = p^2 (1-p) (p + 2 - 2p) = p^2 (1-p) (2 - p)$ Therefore, $$M_X''(0) = \frac{p^2 (1-p) (2 - p)}{p^4} = \frac{(1-p)(2-p)}{p^2}.$$ 5. **Calculate Variance:** $$Var(X) = M_X''(0) - (M_X'(0))^2 = \frac{(1-p)(2-p)}{p^2} - \left(\frac{1-p}{p}\right)^2 = \frac{(1-p)(2-p) - (1-p)^2}{p^2}.$$ Simplify numerator: $$(1-p)(2-p) - (1-p)^2 = (1-p)(2-p - (1-p)) = (1-p)(2-p -1 + p) = (1-p)(1) = 1-p.$$ So, $$Var(X) = \frac{1-p}{p^2}.$$ **Final answers:** - Moment generating function: $$M_X(t) = \frac{p}{1 - (1-p) e^t}.$$ - Mean: $$E[X] = \frac{1-p}{p}.$$ - Variance: $$Var(X) = \frac{1-p}{p^2}.$$