Further Statistics 1

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• For a discrete random variable, first check $\sum p(x)=1$. Then $E(X)=\sum xp(x)$, $E(X^2)=\sum x^2p(x)$ and $\operatorname{Var}(X)=E(X^2)-[E(X)]^2$. For a function, calculate $E[g(X)]=\sum g(x)p(x)$ directly; in general it is not $g(E(X))$.
• Worked example - expectation: if $P(X=0,1,2)=(0.2,0.5,0.3)$, then $E(X)=1.1$, $E(X^2)=1.7$ and $\operatorname{Var}(X)=1.7-1.1^2=0.49$. Also $E[(X-1)^2]=0.2+0+0.3=0.5$.
• If $X\sim\operatorname{Bin}(n,p)$, then $E(X)=np$ and $\operatorname{Var}(X)=np(1-p)$. If $Y\sim\operatorname{Po}(\lambda)$, both mean and variance are $\lambda$. Independent Poisson variables add by adding their parameters.
• Use $\operatorname{Po}(np)$ to approximate $\operatorname{Bin}(n,p)$ when $n$ is large and $p$ is small. Keep the event unchanged: there is no continuity correction in the binomial-to-Poisson step. In Edexcel's CLT questions about a sample mean, model the sample mean directly as continuous unless the question explicitly asks for a correction.
• A geometric variable $X$ counting the trial of the first success has $P(X=x)=(1-p)^{x-1}p$, $E(X)=1/p$ and $\operatorname{Var}(X)=(1-p)/p^2$. A negative binomial variable counting the trial of the $r$th success has $P(X=x)=\binom{x-1}{r-1}p^r(1-p)^{x-r}$, mean $r/p$ and variance $r(1-p)/p^2$.
• For a Poisson-mean or geometric-parameter hypothesis test, write $H_0$ and $H_1$ in the parameter, choose the tail from $H_1$, calculate an exact tail probability or critical region under $H_0$, compare with the significance level, and conclude in the context of the claim.
• The Central Limit Theorem gives $\overline X\mathrel{\dot\sim}N(\mu,\sigma^2/n)$ for a sufficiently large random sample of independent, identically distributed observations with finite variance. Standardise using the standard error $\sigma/\sqrt n$, not $\sigma$.
• For chi-squared goodness of fit, use $\chi^2=\sum\dfrac{(O-E)^2}{E}$ and $df=k-1-m$, where $m$ parameters were estimated from the data. For an $r\times c$ contingency table, $E=\dfrac{(\text{row total})(\text{column total})}{\text{grand total}}$ and $df=(r-1)(c-1)$.
• Worked example - goodness of fit: observed counts $(18,32,50)$ are tested against expected counts $(25,25,50)$. Then $\chi^2=49/25+49/25=3.92$. With $df=2$, $3.92<5.991$, so at $5\%$ there is insufficient evidence against the proposed model.
• A probability generating function is $G_X(t)=E(t^X)$. The standard forms are $(1-p+pt)^n$ for binomial, $e^{\lambda(t-1)}$ for Poisson, $\dfrac{pt}{1-(1-p)t}$ for geometric, and $\left(\dfrac{pt}{1-(1-p)t}\right)^r$ for negative binomial.
• Derivation example - geometric PGF: writing $q=1-p$, $G_X(t)=\sum_{x=1}^{\infty}t^xpq^{x-1}=pt\sum_{r=0}^{\infty}(qt)^r=\dfrac{pt}{1-qt}$. This starts from the definition, substitutes the probability function and sums a geometric series.
• Use $G'_X(1)=E(X)$ and $\operatorname{Var}(X)=G''_X(1)+G'_X(1)-[G'_X(1)]^2$. For independent variables, $G_{X+Y}(t)=G_X(t)G_Y(t)$; coefficients then recover probabilities.
• Worked example - PGFs: for geometric $X$ with $p=0.4$, $G_X(t)=\dfrac{0.4t}{1-0.6t}$, $E(X)=2.5$ and $\operatorname{Var}(X)=3.75$. The sum of two independent copies has PGF $[G_X(t)]^2$, the negative binomial form for the trial of the second success.
• A Type I error rejects a true $H_0$; a Type II error does not reject a false $H_0$. The size is the actual probability of a Type I error. The power function is $\pi(\theta)=P_{\theta}(\text{reject }H_0)$, and at a specified alternative it equals $1-P(\text{Type II error})$.

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