Solution. Throughout, \(M_X(t) = E(e^{tX})\), and the mean and variance come from \[E(X) = M_X'(0), \qquad \operatorname {Var}(X) = M_X''(0)-\left [M_X'(0)\right ]^{2}.\] In every case the range of \(t\) is dictated by where the sum or integral converges.

(a). This is the geometric distribution counting failures before the first success, with \(q = 1-p\). The sum is geometric with ratio \(qe^{t}\): \[M_X(t) = \sum _{x=0}^{\infty }e^{tx}pq^{x} = p\sum _{x=0}^{\infty }\left (qe^{t}\right )^{x} = \frac {p}{1-qe^{t}},\] which converges precisely when \(qe^{t}<1\), that is for \(t < \ln \left (\tfrac {1}{q}\right )\). Differentiating, \[M_X'(t) = \frac {pqe^{t}}{\left (1-qe^{t}\right )^{2}} \implies E(X) = \frac {pq}{(1-q)^{2}} = \frac {pq}{p^{2}} = \frac {q}{p},\] and a second differentiation gives \(M_X''(0) = \frac {q(1+q)}{p^{2}}\), so \[\operatorname {Var}(X) = \frac {q(1+q)}{p^{2}}-\frac {q^{2}}{p^{2}} = \frac {q}{p^{2}}.\]

(b). This is the negative binomial written with a negative upper index. The identity \[\binom {-k}{x}(-1)^{x} = \binom {k+x-1}{x}\] turns it into the familiar form \(\binom {k+x-1}{x}p^{k}q^{x}\): the number of failures before the \(k\)th success. Using the binomial series \((1+z)^{-k} = \sum _{x\geq 0}\binom {-k}{x}z^{x}\) with \(z = -qe^{t}\), \[M_X(t) = p^{k}\sum _{x=0}^{\infty }\binom {-k}{x}\left (-qe^{t}\right )^{x} = p^{k}\left (1-qe^{t}\right )^{-k} = \left (\frac {p}{1-qe^{t}}\right )^{\!k},\] valid for \(t<\ln \left (\tfrac 1q\right )\) as before. This is the \(k\)th power of the m.g.f. in part (a) – exactly what one expects, since a negative binomial is the sum of \(k\) independent geometrics. Hence \[E(X) = \frac {kq}{p}, \qquad \operatorname {Var}(X) = \frac {kq}{p^{2}}.\]

(c). This is the exponential distribution with mean \(\theta \): \[M_X(t) = \int _0^{\infty }e^{tx}\frac {e^{-x/\theta }}{\theta }\,dx = \frac {1}{\theta }\int _0^{\infty }e^{-x\left (\frac {1}{\theta }-t\right )}dx = \frac {1}{\theta }\cdot \frac {1}{\frac {1}{\theta }-t} = \frac {1}{1-\theta t}.\] The integral converges only if \(\tfrac {1}{\theta }-t>0\), so the m.g.f. is valid for \(t<\tfrac {1}{\theta }\). Then \[M_X'(t) = \frac {\theta }{(1-\theta t)^{2}},\qquad M_X''(t) = \frac {2\theta ^{2}}{(1-\theta t)^{3}},\] giving \(E(X) = \theta \) and \(\operatorname {Var}(X) = 2\theta ^{2}-\theta ^{2} = \theta ^{2}\).

(d). This is the gamma distribution with shape \(\alpha \) and scale \(\beta \). \[M_X(t) = \int _0^{\infty }e^{tx}\frac {x^{\alpha -1}e^{-x/\beta }}{\beta ^{\alpha }\Gamma (\alpha )}dx = \frac {1}{\beta ^{\alpha }\Gamma (\alpha )}\int _0^{\infty }x^{\alpha -1} e^{-x\left (\frac {1}{\beta }-t\right )}dx.\] The remaining integral is a gamma integral: substituting \(u = x\left (\tfrac {1}{\beta }-t\right )\) gives \(\Gamma (\alpha )\left (\tfrac {1}{\beta }-t\right )^{-\alpha }\), so \[M_X(t) = \frac {\left (\frac {1}{\beta }-t\right )^{-\alpha }}{\beta ^{\alpha }} = (1-\beta t)^{-\alpha }, \qquad t<\frac {1}{\beta }.\] Then \[M_X'(t) = \alpha \beta (1-\beta t)^{-\alpha -1},\qquad M_X''(t) = \alpha (\alpha +1)\beta ^{2}(1-\beta t)^{-\alpha -2},\] so \(E(X) = \alpha \beta \) and \[\operatorname {Var}(X) = \alpha (\alpha +1)\beta ^{2}-\alpha ^{2}\beta ^{2} = \alpha \beta ^{2}.\] Setting \(\alpha = 1\) recovers part (c), as it must: the exponential is a gamma with shape 1.

Solution. (a). Let \(X\) be the number of heads, so \(X\sim B\left (5,\tfrac 12\right )\) and \(P(X=x) = \binom {5}{x}\left (\tfrac 12\right )^{5} = \binom {5}{x}/32\). The player pays $4 and receives $\(X\), so the net winnings are \(N = X-4\), taking values \(-4\) to \(1\):

These sum to \(\tfrac {32}{32}=1\).

(b). Subtracting a constant shifts the mean and leaves the variance alone, so there is no need to work from the table: \[E(N) = E(X)-4 = 5\left (\tfrac 12\right )-4 = 2.5-4 = -1.5,\] \[\operatorname {Var}(N) = \operatorname {Var}(X) = npq = 5\left (\tfrac 12\right )\left (\tfrac 12\right ) = 1.25.\] The game is not fair: on average the player loses $1.50 per game. To make it fair the stake would have to be $2.50, the expected number of heads.

Solution. (a). A finite sum of the form \(\sum p_i e^{tx_i}\) is the m.g.f. of a discrete variable, and the coefficients are the probabilities. Reading it off: \(X\) takes the value \(-1\) with probability \(\tfrac 16\), \(-2\) with probability \(\tfrac 12\), and \(1\) with probability \(\tfrac 13\). (These sum to 1, as they must.) So there is no need to differentiate at all: \begin {align*} E(X) &= (-1)\tfrac 16+(-2)\tfrac 12+(1)\tfrac 13 = -\tfrac 16-1+\tfrac 13 = -\tfrac 56,\\ E(X^2) &= (1)\tfrac 16+(4)\tfrac 12+(1)\tfrac 13 = \tfrac 52,\\ \operatorname {Var}(X) &= \tfrac 52-\left (-\tfrac 56\right )^{2} = \tfrac {90-25}{36} = \frac {65}{36} \approx 1.806. \end {align*}

(b). Expanding each exponential as \(e^{u} = \sum u^n/n!\) and collecting powers of \(t\) – equivalently, using \(M_X(t) = \sum E(X^n)t^n/n!\) with \(E(X^3) = -\tfrac {23}{6}\) and \(E(X^4) = \tfrac {17}{2}\): \[M_X(t) = 1-\frac {5}{6}t+\frac {5}{4}t^{2}-\frac {23}{36}t^{3}+\frac {17}{48}t^{4}+\cdots \] The coefficient of \(t^n\) is \(E(X^n)/n!\), which is what makes the expansion a way of reading off moments.

Solution. (a). The probabilities must sum to 1. The series starts at \(x=1\), so \[\sum _{x=1}^{\infty }k\left (\frac {1}{2}\right )^{x} = k\cdot \frac {\frac 12}{1-\frac 12} = k = 1 \implies k = 1.\] So \(f(x) = \left (\tfrac 12\right )^{x}\): a geometric distribution on \(\{1,2,3,\ldots \}\) with \(p = \tfrac 12\).

(b). \[M_X(t) = \sum _{x=1}^{\infty }e^{tx}\left (\frac {1}{2}\right )^{x} = \sum _{x=1}^{\infty }\left (\frac {e^{t}}{2}\right )^{x} = \frac {e^{t}/2}{1-e^{t}/2} = \frac {e^{t}}{2-e^{t}},\] valid when \(e^{t}/2<1\), that is \(t<\ln 2\).

(c). Differentiating by the quotient rule, \[M_X'(t) = \frac {2e^{t}}{\left (2-e^{t}\right )^{2}} \implies E(X) = \frac {2}{1} = 2,\] and a second differentiation gives \(M_X''(0) = 6\), so \[\operatorname {Var}(X) = 6-2^{2} = 2.\] These agree with the standard geometric results \(E(X)=1/p = 2\) and \(\operatorname {Var}(X) = q/p^{2} = 2\).

(d). For a linear transformation \(Y = aX+b\), \(M_Y(t) = e^{bt}M_X(at)\). With \(a=2\), \(b=-5\): \[M_Y(t) = e^{-5t}M_X(2t) = e^{-5t}\cdot \frac {e^{2t}}{2-e^{2t}} = \frac {e^{-3t}}{2-e^{2t}},\] valid for \(2t<\ln 2\), that is \(t<\tfrac {1}{2}\ln 2\).

(e). Either differentiate \(M_Y\), or use the transformation rules directly: \[E(Y) = 2E(X)-5 = 4-5 = -1, \qquad \operatorname {Var}(Y) = 2^{2}\operatorname {Var}(X) = 4(2) = 8.\] The shift \(-5\) moves the mean but not the variance; the factor 2 scales the variance by \(2^{2}\).

Solution. This is the Laplace (double exponential) density; note it can be written compactly as \(f(x) = \tfrac 12 e^{-|x|}\), which makes its symmetry about zero obvious.

(a). Split the integral at the origin: \begin {align*} M_X(t) &= \int _{-\infty }^{0}e^{tx}\cdot \frac {1}{2}e^{x}\,dx + \int _{0}^{\infty }e^{tx}\cdot \frac {1}{2}e^{-x}\,dx\\ &= \frac {1}{2}\int _{-\infty }^{0}e^{(1+t)x}\,dx + \frac {1}{2}\int _{0}^{\infty }e^{-(1-t)x}\,dx\\ &= \frac {1}{2}\cdot \frac {1}{1+t}+\frac {1}{2}\cdot \frac {1}{1-t} = \frac {1}{2}\cdot \frac {(1-t)+(1+t)}{(1+t)(1-t)} = \frac {1}{1-t^{2}}. \end {align*}

The first integral needs \(1+t>0\) and the second needs \(1-t>0\), so both hold precisely when \(|t|<1\).

(b). By symmetry \(E(X)=0\); formally, \(M_X'(t) = 2t\left (1-t^{2}\right )^{-2}\) vanishes at \(t=0\). Differentiating again, \[M_X''(t) = 2\left (1-t^{2}\right )^{-2}+8t^{2}\left (1-t^{2}\right )^{-3} \implies E(X^{2}) = M_X''(0) = 2,\] so \(\operatorname {Var}(X) = 2-0^{2} = 2\).