Probability & Statistics

The three judges give guilty verdicts with probabilities $p$, $p/3$, and $p^2$ independently. Since $X$ counts the total number of guilty verdicts, linearity of expectation gives the result immediately without computing the full distribution:$$\mathrm{E}(X)=p+\frac{p}{3}+p^2=\frac{4p}{3}+p^2.$$Setting this equal to $4/3$:$$\frac{4p}{3}+p^2=\frac{4}{3}⟹3p^2+4p-4=0⟹(3p-2)(p+2)=0.$$Since $0<p<1$, we get $p=\frac{2}{3}$.

Finding t.

With $p=2/3$, compute the probabilities $P(X=k)$ for $k=0,1,2,3$. The King pronounces guilty when $X\ge 2$, so the key probabilities are:$$P(X=2)=p\cdot \frac{p}{3}\cdot (1-p^2)+p\cdot (1-\frac{p}{3})\cdot p^2+(1-p)\cdot \frac{p}{3}\cdot p^2=\frac{1}{3}{p}^2(1+4p-3p^2),$$$$P(X=3)=p\cdot \frac{p}{3}\cdot p^2=\frac{p^4}{3}.$$With $p=2/3$: $P(X=3)=\frac{(2/3{)}^4}{3}=\frac{16}{243}$ and $P(X=2)=\frac{1}{3}\cdot \frac{4}{9}\cdot \left(1+\frac{8}{3}-\frac{4}{3}\right)=\frac{84}{243}$.

So $P(\text{King says guilty})=P(X\ge 2)=\frac{100}{243}$, and $P(\text{King says not guilty})=\frac{143}{243}$.

A correct pronouncement occurs in exactly two scenarios: the defendant is guilty and $X\ge 2$, or the defendant is not guilty and $X\le 1$. Using $t=P(\text{defendant is guilty})$:$$P(\text{correct})=t\cdot \frac{100}{243}+(1-t)\cdot \frac{143}{243}=\frac{143-43t}{243}.$$Setting this equal to $1/2$:$$143-43t=\frac{243}{2}⟹43t=143-\frac{243}{2}=\frac{43}{2}⟹t=\frac{1}{2}.$$This elegant result has a neat interpretation: when $t=1/2$, the defendant is equally likely to be guilty or not guilty going in. In that situation, no matter what the judges decide, the King is right exactly half the time — the system provides no useful information whatsoever.

We have $n$ pieces of string with $2n$ free ends in total. At each step we pick two ends uniformly at random and tie them together.

Expected rings at the first step.

At the very first step there are $2n$ free ends. We pick one end; there are $2n-1$ remaining ends. The only way to create a ring is if we pick the other end of the same piece of string — and there is exactly one such end among the $2n-1$ remaining. So$$P(\text{ring at step 1})=\frac{1}{2n-1}.$$The expected number of rings created at step 1 is therefore $\frac{1}{2n-1}$.

Expected total rings.

The key observation is that the number of free ends decreases by exactly 2 at each step, regardless of whether a ring was formed or not. After step $k$, there are $2n-2k$ free ends. At the $(k+1)$th step, the probability of forming a ring is $\frac{1}{2(n-k)-1}$.

Letting $k$ run from $0$ to $n-1$, the expected total number of rings is$$E=\sum _{k=0}^{n-1}\frac{1}{2(n-k)-1}=\frac{1}{2n-1}+\frac{1}{2n-3}+\cdots +\frac{1}{3}+\frac{1}{1}.$$This is the sum of reciprocals of all odd numbers from $1$ to $2n-1$.

Variance.

At each step the number of rings created is a Bernoulli random variable: at step $k+1$ it equals $1$ with probability $p_k=\frac{1}{2(n-k)-1}$ and $0$ otherwise. Its variance is $p_k(1-p_k)$.

Since the outcomes at different steps are independent (the composition of the strings is randomised afresh at each step in distribution), the total variance is$$\mathrm{Var}=\sum _{k=0}^{n-1}{p}_k(1-p_k)=\sum _{k=0}^{n-1}\frac{1}{2(n-k)-1}\left(1-\frac{1}{2(n-k)-1}\right)=\sum _{j=0}^{n-1}\frac{2(n-1-j)}{(2j+1{)}^2}$$or equivalently$$\mathrm{Var}=\frac{2(n-1)}{(2n-1{)}^2}+\frac{2(n-2)}{(2n-3{)}^2}+\cdots +\frac{2}{3^2}.$$Approximation for $n=40000$.

We need$$E=1+\frac{1}{3}+\frac{1}{5}+\cdots +\frac{1}{79999}.$$Write the sum of reciprocals of odd integers up to $2n-1$ in terms of the harmonic series:$$E=\sum _{k=1}^{2n}\frac{1}{k}-\sum _{k=1}^n\frac{1}{2k}=\sum _{k=1}^{2n}\frac{1}{k}-\frac{1}{2}\sum _{k=1}^n\frac{1}{k}\approx \ln (2n)-\frac{1}{2}\ln n=\ln 2+\frac{1}{2}\ln n.$$For $n=40000$: $\frac{1}{2}\ln (40000)=\frac{1}{2}\ln (20^2)=\ln 20\approx 3$. Also $\ln 2\approx 0.69$. Alternatively, note $2n=80000=2\times 40000$, so$$E\approx \ln (80000)-\frac{1}{2}\ln (40000)=\ln (80000)-\ln (200)=\ln \left(\frac{80000}{200}\right)=\ln 400.$$Using a cleaner route: $\ln (80000)-\frac{1}{2}\ln (40000)=\ln (80000)-\ln \sqrt{40000}$. Since $\sqrt{40000}=200$ and $80000/200=400=20^2$, we get $E\approx \ln (20^2)=2\ln 20\approx 6$.

The expected number of rings when $n=40000$ is approximately $6$.

Skewness measures the asymmetry of a distribution — whether values are spread out more to the left or to the right of the mean.

Part (i). We expand $(X-\mu {)}^3$ inside the expectation:$$E((X-\mu {)}^3)=E(X^3)-3\mu E(X^2)+3{\mu }^{2}E(X)-{\mu }^3.$$Since $E(X)=\mu$ and $E(X^2)={\sigma }^2+{\mu }^2$, substituting gives$$E((X-\mu {)}^3)=E(X^3)-3\mu ({\sigma }^2+{\mu }^2)+3{\mu }^3-{\mu }^3=E(X^3)-3\mu {\sigma }^2-{\mu }^3.$$Dividing through by ${\sigma }^3$ yields the required formula for $\gamma$.

Computing $\gamma$ for $f(x)=2x$ on $[0,1]$. The moments are$$E(X)={\int }_0^{1}2x^{2}dx=\frac{2}{3},E(X^2)={\int }_0^{1}2x^{3}dx=\frac{1}{2},E(X^3)={\int }_0^{1}2x^{4}dx=\frac{2}{5}.$$So $\mu =\frac{2}{3}$, ${\sigma }^2=\frac{1}{2}-\frac{4}{9}=\frac{1}{18}$, and ${\sigma }^3=(\frac{1}{18}{)}^{3/2}=\frac{1}{18\sqrt{18}}$. Substituting:$$\gamma =\frac{\frac{2}{5}-3\cdot \frac{2}{3}\cdot \frac{1}{18}-\frac{8}{27}}{\frac{1}{18\sqrt{18}}}=\frac{\frac{2}{5}-\frac{1}{9}-\frac{8}{27}}{\frac{1}{18\sqrt{18}}}.$$The numerator: a common denominator of 135 gives $\frac{54-15-40}{135}=-\frac{1}{135}$. The denominator equals $\frac{\sqrt{2}}{18\cdot 6}=\frac{\sqrt{2}}{108}$ (since $\sqrt{18}=3\sqrt{2}$, so $18\sqrt{18}=54\sqrt{2}$, giving ${\sigma }^3=\frac{1}{54\sqrt{2}}$). Hence$$\gamma =-\frac{1}{135}\times 54\sqrt{2}=-\frac{54\sqrt{2}}{135}=-\frac{2\sqrt{2}}{5}.$$Part (ii). The CDF is $F(x)=x^2$ for $x\in [0,1]$, so $F^{-1}(p)=\sqrt{p}$. The three quantiles are:$$F^{-1}\left(\frac{9}{10}\right)=\sqrt{\frac{9}{10}}=\frac{3}{\sqrt{10}},F^{-1}\left(\frac{1}{2}\right)=\frac{1}{\sqrt{2}},F^{-1}\left(\frac{1}{10}\right)=\frac{1}{\sqrt{10}}.$$The numerator of $D$ is $\frac{3}{\sqrt{10}}-\frac{2}{\sqrt{2}}+\frac{1}{\sqrt{10}}=\frac{4}{\sqrt{10}}-\frac{2}{\sqrt{2}}$, and the denominator is $\frac{3}{\sqrt{10}}-\frac{1}{\sqrt{10}}=\frac{2}{\sqrt{10}}$. So$$D=\frac{\frac{4}{\sqrt{10}}-\frac{2}{\sqrt{2}}}{\frac{2}{\sqrt{10}}}=2-\frac{\sqrt{10}}{\sqrt{2}}=2-\sqrt{5}.$$Pearson skewness. The median satisfies $M^2=\frac{1}{2}$, so $M=\frac{1}{\sqrt{2}}$. With $\mu =\frac{2}{3}$ and $\sigma =\frac{1}{\sqrt{18}}=\frac{1}{3\sqrt{2}}$:$$P=\frac{3(\frac{2}{3}-\frac{1}{\sqrt{2}})}{\frac{1}{3\sqrt{2}}}=9\sqrt{2}\left(\frac{2}{3}-\frac{1}{\sqrt{2}}\right)=6\sqrt{2}-9.$$Showing $D>P>\gamma$. All three values are negative, so we work numerically: $D=2-\sqrt{5}\approx -0.236$, $P=6\sqrt{2}-9\approx -0.485$, $\gamma =-\frac{2\sqrt{2}}{5}\approx -0.566$.

For $D>P$: we need $2-\sqrt{5}>6\sqrt{2}-9$, i.e. $11-\sqrt{5}>6\sqrt{2}$. Squaring (both sides positive): $121-22\sqrt{5}+5>72$, so $54>22\sqrt{5}$, i.e. $27>11\sqrt{5}$. Squaring again: $729>605$. True.

For $P>\gamma$: $6\sqrt{2}-9>-\frac{2\sqrt{2}}{5}$, i.e. $6\sqrt{2}+\frac{2\sqrt{2}}{5}>9$, so $\frac{32\sqrt{2}}{5}>9$, i.e. $32\sqrt{2}>45$. Squaring: $2048>2025$. True.