{ "deckName": "Exam P — Expectation, Variance & MGFs", "examCode": "Exam P", "cards": [ { "front": "Define the expected value $E[X]$ for a discrete and for a continuous random variable.", "back": "For a discrete $X$ with pmf $p(x)$: $E[X]=\\sum_{x} x\\,p(x)$.\nFor a continuous $X$ with pdf $f(x)$: $E[X]=\\int_{-\\infty}^{\\infty} x\\,f(x)\\,dx$.\n$E[X]$ exists only if the sum/integral converges absolutely.", "tag": "Expectation" }, { "front": "State the Law of the Unconscious Statistician (LOTUS) for $E[g(X)]$.", "back": "You do **not** need the distribution of $Y=g(X)$ to find its mean. Use the distribution of $X$ directly:\nDiscrete: $E[g(X)]=\\sum_{x} g(x)\\,p(x)$.\nContinuous: $E[g(X)]=\\int_{-\\infty}^{\\infty} g(x)\\,f(x)\\,dx$.", "tag": "Expectation" }, { "front": "Is $E[g(X)]=g(E[X])$ in general? Give the rule.", "back": "No. Equality holds when $g$ is affine, $g(x)=ax+b$ — more generally when $g$ is affine on the support of $X$, or when $X$ is degenerate (constant). For a convex $g$, **Jensen's inequality** gives $E[g(X)]\\geq g(E[X])$ (e.g. $E[X^{2}]\\geq (E[X])^{2}$), strict when $g$ is strictly convex and $X$ is nondegenerate; for concave $g$ the inequality reverses.", "tag": "Expectation" }, { "front": "For a nonnegative continuous random variable, express $E[X]$ using the survival function $S(x)=1-F(x)$.", "back": "$E[X]=\\int_{0}^{\\infty} S(x)\\,dx=\\int_{0}^{\\infty}\\big(1-F(x)\\big)\\,dx$.\nFor a nonnegative integer-valued $X$: $E[X]=\\sum_{k=1}^{\\infty}\\Pr(X\\geq k)$. This 'tail formula' is very efficient for deductible/limit problems.", "tag": "Expectation" }, { "front": "$X$ takes values $0,1,2,3$ with probabilities $0.4,0.3,0.2,0.1$. Find $E[X]$ and $E[X^{2}]$.", "back": "$E[X]=0(0.4)+1(0.3)+2(0.2)+3(0.1)=0+0.3+0.4+0.3=1.0$.\n$E[X^{2}]=0(0.4)+1(0.3)+4(0.2)+9(0.1)=0+0.3+0.8+0.9=2.0$.\nSo $E[X]=1.0$ and $E[X^{2}]=2.0$.", "tag": "Expectation" }, { "front": "$X$ has pdf $f(x)=2x$ on $00$, a minimum). The minimum value is $E[X^{2}]-(E[X])^{2}=\\operatorname{Var}(X)$. So the mean is the least-squares center, and the variance is that minimal mean-squared deviation.", "tag": "Variance" }, { "front": "How do $E$ and $\\operatorname{Var}$ behave under the linear transform $aX+b$?", "back": "$E[aX+b]=a\\,E[X]+b$.\n$\\operatorname{Var}(aX+b)=a^{2}\\operatorname{Var}(X)$, so $\\operatorname{SD}(aX+b)=|a|\\,\\sigma$.\nA shift $b$ moves the mean but never changes the spread; only the scale $a$ does, and it enters squared.", "tag": "Linear transforms" }, { "front": "$X$ has $E[X]=20$ and $\\operatorname{Var}(X)=9$. Let $Y=3X-5$. Find $E[Y]$, $\\operatorname{Var}(Y)$, and $\\operatorname{SD}(Y)$.", "back": "$E[Y]=3E[X]-5=3(20)-5=55$.\n$\\operatorname{Var}(Y)=3^{2}\\operatorname{Var}(X)=9(9)=81$.\n$\\operatorname{SD}(Y)=\\sqrt{81}=9=|3|\\cdot\\sqrt{9}$.", "tag": "Linear transforms" }, { "front": "Convert a Fahrenheit temperature $F$ to Celsius via $C=\\frac{5}{9}(F-32)$. If the standard deviation of $F$ is $\\operatorname{SD}(F)=18$, what is $\\operatorname{SD}(C)$?", "back": "Only the multiplicative factor $\\frac{5}{9}$ affects spread; the shift $-32\\cdot\\frac{5}{9}$ does not.\n$\\operatorname{SD}(C)=\\big|\\frac{5}{9}\\big|\\cdot\\operatorname{SD}(F)=\\frac{5}{9}\\times 18=10$.\nSo the standard deviation of the Celsius reading is $10$ degrees.", "tag": "Linear transforms" }, { "front": "Independent $X,Y$ have $\\operatorname{Var}(X)=4,\\operatorname{Var}(Y)=9$. Find $\\operatorname{Var}(2X-Y+5)$.", "back": "The constant $+5$ does not affect variance.\n$\\operatorname{Var}(2X-Y)=2^{2}\\operatorname{Var}(X)+(-1)^{2}\\operatorname{Var}(Y)=4(4)+1(9)=16+9=25$.\nSo $\\operatorname{Var}(2X-Y+5)=25$. (Independence: variances add for both sum and difference.)", "tag": "Linear transforms" }, { "front": "Independent $X_{1},\\dots,X_{25}$ are i.i.d. with mean $\\mu=8$ and variance $\\sigma^{2}=4$. Find $E[\\bar{X}]$ and $\\operatorname{Var}(\\bar{X})$ for the sample mean $\\bar{X}$.", "back": "$E[\\bar{X}]=\\mu=8$ (the sample mean is unbiased for the mean).\n$\\operatorname{Var}(\\bar{X})=\\frac{\\sigma^{2}}{n}=\\frac{4}{25}=0.16$, so $\\operatorname{SD}(\\bar{X})=\\frac{\\sigma}{\\sqrt{n}}=\\frac{2}{5}=0.4$.\nAveraging shrinks variance by a factor of $n$.", "tag": "Linear transforms" }, { "front": "Define the $n$-th raw moment and the $n$-th central moment of $X$.", "back": "Raw (about the origin): $\\mu_{n}'=E[X^{n}]$.\nCentral (about the mean $\\mu=E[X]$): $\\mu_{n}=E\\big[(X-\\mu)^{n}\\big]$.\nNote $\\mu_{1}'=\\mu$, $\\mu_{1}=0$, and $\\mu_{2}=\\operatorname{Var}(X)$.", "tag": "Moments" }, { "front": "$X$ has $E[X]=5$ and $\\operatorname{Var}(X)=9$. Find $E[X^{2}]$ and $E[(2X-3)^{2}]$.", "back": "$E[X^{2}]=\\operatorname{Var}(X)+(E[X])^{2}=9+25=34$.\nLet $Y=2X-3$: $E[Y]=2(5)-3=7$, $\\operatorname{Var}(Y)=2^{2}(9)=36$.\n$E[Y^{2}]=\\operatorname{Var}(Y)+(E[Y])^{2}=36+49=85$.\nSo $E[(2X-3)^{2}]=85$.", "tag": "Moments" }, { "front": "$X$ has pdf $f(x)=3x^{2}$ on $01$ (a Pareto-type tail). Find $E[X]$ and explain whether $\\operatorname{Var}(X)$ is finite.", "back": "$E[X]=\\int_{1}^{\\infty}x\\cdot\\frac{2}{x^{3}}dx=\\int_{1}^{\\infty}2x^{-2}dx=\\big[-2x^{-1}\\big]_{1}^{\\infty}=2$.\n$E[X^{2}]=\\int_{1}^{\\infty}x^{2}\\cdot\\frac{2}{x^{3}}dx=\\int_{1}^{\\infty}2x^{-1}dx=\\big[2\\ln x\\big]_{1}^{\\infty}=\\infty$.\nSince $E[X^{2}]$ diverges, $\\operatorname{Var}(X)$ is **infinite** — a hallmark of heavy tails (and no MGF).", "tag": "Moments" }, { "front": "A risk pays benefit $0$ with probability $0.9$ and a $\\operatorname{Uniform}(0,1000)$ amount with probability $0.1$. Find $E[X]$ and $\\operatorname{Var}(X)$ of this mixed payment.", "back": "Let $U\\sim\\operatorname{Uniform}(0,1000)$: $E[U]=500$, $E[U^{2}]=\\frac{1000^{2}}{3}=333{,}333.33$.\n$E[X]=0.9(0)+0.1(500)=50$.\n$E[X^{2}]=0.9(0)+0.1(333{,}333.33)=33{,}333.33$.\n$\\operatorname{Var}(X)=33{,}333.33-50^{2}=33{,}333.33-2500=30{,}833.33$.", "tag": "Moments" }, { "front": "Define skewness and state what its sign tells you about a distribution.", "back": "Skewness $\\gamma_{1}=\\frac{E[(X-\\mu)^{3}]}{\\sigma^{3}}=\\frac{\\mu_{3}}{\\sigma^{3}}$.\n$\\gamma_{1}>0$: right-skewed (long right tail, mean above median — e.g. the exponential). $\\gamma_{1}<0$: left-skewed. $\\gamma_{1}=0$ for any symmetric distribution.", "tag": "Shape" }, { "front": "Define kurtosis and excess kurtosis, and give the normal benchmark.", "back": "Kurtosis $\\frac{E[(X-\\mu)^{4}]}{\\sigma^{4}}=\\frac{\\mu_{4}}{\\sigma^{4}}$. Excess kurtosis subtracts $3$. A normal distribution has kurtosis $3$ (excess $0$). Excess $>0$ means heavier tails / sharper peak (leptokurtic); excess $<0$ means lighter tails (platykurtic).", "tag": "Shape" }, { "front": "Define the coefficient of variation (CV) and state when it is most useful.", "back": "$\\operatorname{CV}=\\frac{\\operatorname{SD}(X)}{E[X]}=\\frac{\\sigma}{\\mu}$ (often as a percentage), defined for $\\mu\\neq 0$, usually for positive $X$.\nIt is a **scale-free** measure of relative dispersion, letting you compare variability across variables with different units or magnitudes.", "tag": "Coefficient of variation" }, { "front": "A loss has $E[X]=2{,}000$ and $\\operatorname{Var}(X)=2{,}560{,}000$. Find its coefficient of variation.", "back": "$\\operatorname{SD}(X)=\\sqrt{2{,}560{,}000}=1{,}600$.\n$\\operatorname{CV}=\\frac{\\operatorname{SD}(X)}{E[X]}=\\frac{1{,}600}{2{,}000}=0.8$ (i.e. $80\\%$).", "tag": "Coefficient of variation" }, { "front": "What is the coefficient of variation of any exponential distribution, and why is it constant?", "back": "For an exponential with mean $\\theta$: $\\operatorname{SD}=\\theta$ and $E[X]=\\theta$, so $\\operatorname{CV}=\\theta/\\theta=1$.\nThe CV is scale-free, and changing $\\theta$ is just a rescaling $X\\to cX$, which leaves $\\operatorname{SD}/\\operatorname{mean}$ unchanged. So every exponential has $\\operatorname{CV}=1$.", "tag": "Coefficient of variation" }, { "front": "An aggregate is $S=X_{1}+X_{2}+X_{3}$ with the $X_{i}$ independent, $E[X_{i}]=100$, $\\operatorname{Var}(X_{i})=400$. Find $E[S]$, $\\operatorname{Var}(S)$, and $\\operatorname{CV}(S)$.", "back": "$E[S]=3\\times100=300$.\n$\\operatorname{Var}(S)=3\\times400=1200$ (independence: variances add).\n$\\operatorname{SD}(S)=\\sqrt{1200}\\approx 34.64$, so $\\operatorname{CV}(S)=\\frac{34.64}{300}\\approx 0.115$.\nThe CV of the sum ($\\approx0.115$) is smaller than each component's CV ($20/100=0.2$) — pooling diversifies.", "tag": "Coefficient of variation" }, { "front": "Define the moment generating function $M_{X}(t)$ and state where it must exist.", "back": "$M_{X}(t)=E\\big[e^{tX}\\big]$.\nDiscrete: $\\sum_{x} e^{tx}p(x)$; continuous: $\\int_{-\\infty}^{\\infty} e^{tx}f(x)\\,dx$.\nIt must be finite for $t$ in some open interval containing $0$ for the MGF to be useful. Always $M_{X}(0)=1$.", "tag": "MGF" }, { "front": "How do you recover the moments $E[X^{n}]$ from $M_{X}(t)$?", "back": "Differentiate $n$ times and evaluate at $0$:\n$E[X^{n}]=M_{X}^{(n)}(0)=\\frac{d^{n}}{dt^{n}}M_{X}(t)\\Big|_{t=0}$.\nSo $E[X]=M_{X}'(0)$ and $E[X^{2}]=M_{X}''(0)$; this is why it 'generates' moments.", "tag": "MGF" }, { "front": "Given $M_{X}(t)=e^{3t+8t^{2}}$, identify $E[X]$ and $\\operatorname{Var}(X)$.", "back": "This is the normal MGF $\\exp(\\mu t+\\frac{1}{2}\\sigma^{2}t^{2})$. Matching exponents: $\\mu=3$ and $\\frac{1}{2}\\sigma^{2}=8\\Rightarrow\\sigma^{2}=16$.\nSo $E[X]=3$ and $\\operatorname{Var}(X)=16$ (with $\\operatorname{SD}=4$).", "tag": "MGF" }, { "front": "From $M_{X}(t)=(1-2t)^{-3}$ for $t<\\frac{1}{2}$, compute $E[X]$ and $\\operatorname{Var}(X)$ by differentiation.", "back": "$M_{X}'(t)=6(1-2t)^{-4}\\Rightarrow E[X]=M_{X}'(0)=6$.\n$M_{X}''(t)=48(1-2t)^{-5}\\Rightarrow E[X^{2}]=M_{X}''(0)=48$.\n$\\operatorname{Var}(X)=48-6^{2}=48-36=12$.\n(This is a gamma with $\\alpha=3,\\theta=2$: mean $\\alpha\\theta=6$, variance $\\alpha\\theta^{2}=12$.)", "tag": "MGF" }, { "front": "State the MGF transformation rule for $Y=aX+b$ and apply it: if $X$ is exponential with $M_{X}(t)=\\frac{1}{1-2t}$, find $M_{Y}(t)$ for $Y=3X+1$.", "back": "Rule: $M_{aX+b}(t)=e^{bt}\\,M_{X}(at)$ (the shift gives $e^{bt}$, the scale rescales the argument).\nWith $a=3,b=1$: $M_{Y}(t)=e^{t}\\,M_{X}(3t)=e^{t}\\cdot\\frac{1}{1-2(3t)}=\\frac{e^{t}}{1-6t}$, valid for $t<\\frac{1}{6}$.", "tag": "MGF" }, { "front": "State the uniqueness property of MGFs and why it matters on Exam P.", "back": "If two random variables have MGFs that are equal (and finite) on an open interval around $0$, they have the **same distribution**. So an MGF determines the distribution uniquely. This lets you 'recognize the distribution from its MGF' and read off its mean/variance from the standard form.", "tag": "MGF" }, { "front": "Let $X\\sim\\operatorname{Poisson}(2)$ and $Y\\sim\\operatorname{Poisson}(3)$ be independent. Use MGFs to identify the distribution of $X+Y$.", "back": "For independent sums MGFs multiply, $M_{X+Y}(t)=M_{X}(t)M_{Y}(t)$. Poisson MGF: $\\exp[\\lambda(e^{t}-1)]$.\n$M_{X+Y}(t)=e^{2(e^{t}-1)}\\,e^{3(e^{t}-1)}=e^{5(e^{t}-1)}$.\nBy uniqueness this is $\\operatorname{Poisson}(5)$. Independent Poissons add: $\\lambda=2+3=5$.", "tag": "MGF" }, { "front": "Given the MGF $M_{X}(t)=\\frac{0.4e^{t}}{1-0.6e^{t}}$ for $e^{t}<\\frac{1}{0.6}$, identify the distribution and its mean and variance.", "back": "This matches the geometric MGF $\\frac{pe^{t}}{1-qe^{t}}$ with $p=0.4,\\,q=0.6$ on support $\\{1,2,\\dots\\}$.\nBy uniqueness $X$ is geometric (trials to first success): $E[X]=\\frac{1}{p}=\\frac{1}{0.4}=2.5$ and $\\operatorname{Var}(X)=\\frac{q}{p^{2}}=\\frac{0.6}{0.16}=3.75$.", "tag": "MGF" }, { "front": "Define the probability generating function (PGF) $G_{X}(s)$ and say which variables it applies to.", "back": "For a **nonnegative integer-valued** $X$: $G_{X}(s)=E\\big[s^{X}\\big]=\\sum_{k=0}^{\\infty} s^{k}\\Pr(X=k)$.\nIt converges at least for $|s|\\leq 1$, with $G_{X}(1)=1$. It relates to the MGF by $G_{X}(s)=M_{X}(\\ln s)$.", "tag": "PGF" }, { "front": "How do you recover $\\Pr(X=k)$ and the factorial moments from the PGF $G_{X}(s)$?", "back": "$\\Pr(X=k)=\\frac{G_{X}^{(k)}(0)}{k!}$ (the $k$-th Taylor coefficient).\nFactorial moments at $s=1$: $G_{X}'(1)=E[X]$ and $G_{X}''(1)=E[X(X-1)]$, so $\\operatorname{Var}(X)=G''_{X}(1)+G'_{X}(1)-[G'_{X}(1)]^{2}$.", "tag": "PGF" }, { "front": "A count has PGF $G_{X}(s)=e^{4(s-1)}$. Find $\\Pr(X=0)$, $E[X]$, and $\\operatorname{Var}(X)$.", "back": "$\\Pr(X=0)=G_{X}(0)=e^{4(0-1)}=e^{-4}$.\n$G_{X}'(s)=4e^{4(s-1)}\\Rightarrow E[X]=G_{X}'(1)=4$.\n$G_{X}''(s)=16e^{4(s-1)}\\Rightarrow E[X(X-1)]=G_{X}''(1)=16$.\n$\\operatorname{Var}(X)=16+4-4^{2}=4$. (This is $\\operatorname{Poisson}(4)$: mean $=$ variance $=4$.)", "tag": "PGF" }, { "front": "State Markov's inequality, its key requirement, and apply it: claim sizes are nonnegative with mean $E[X]=400$ — bound $\\Pr(X\\geq 2000)$.", "back": "Markov (requires $X\\geq 0$): $\\Pr(X\\geq a)\\leq \\frac{E[X]}{a}$ for $a>0$.\nWith $a=2000$: $\\Pr(X\\geq 2000)\\leq \\frac{400}{2000}=0.2$.\nAt most $20\\%$ of claims can be $2000$ or larger. The bound uses only the mean and is usually loose.", "tag": "Inequalities" }, { "front": "State Chebyshev's inequality, then apply it: a variable has $\\mu=50$ and $\\sigma=5$ — bound $\\Pr(40