Continuous Random Variables

(a) Impose total area one. The density vanishes outside $[0,2]$, so

$${\int }_0^{2}cx(2-x)dx=c{\int }_0^2(2x-x^2)dx=c{\left[x^2-\frac{x^3}{3}\right]}_0^2=c\left(4-\frac{8}{3}\right)=\frac{4c}{3}=1,$$

giving $c=\frac{3}{4}$. Non-negativity: on $[0,2]$ both factors $x$ and $2-x$ are $\ge 0$, so $f(x)\ge 0$ there, and $f=0$ elsewhere. Both conditions hold, so $f(x)=\frac{3}{4}x(2-x)$ is a genuine density.

(b) Integrate from the left. For $x<0$ nothing has accumulated, so $F(x)=0$. For $0\le x\le 2$,

$$F(x)={\int }_0^x\frac{3}{4}(2t-t^2)dt=\frac{3}{4}\left(x^2-\frac{x^3}{3}\right)=\frac{3x^2-x^3}{4}.$$

For $x>2$ all the mass is to the left, so $F(x)=1$; as a check, $F(2)=\frac{12-8}{4}=1$, so the pieces join continuously. Differentiating on $(0,2)$: $F^{\prime }(x)=\frac{6x-3x^2}{4}=\frac{3}{4}x(2-x)=f(x)$, as the theorem that density and distribution determine each other requires.

(c) Subtract. First the two values of $F$:

$$F\left(\frac{3}{2}\right)=\frac{3\cdot \frac{9}{4}-\frac{27}{8}}{4}=\frac{\frac{27}{4}-\frac{27}{8}}{4}=\frac{27/8}{4}=\frac{27}{32},F\left(\frac{1}{2}\right)=\frac{\frac{3}{4}-\frac{1}{8}}{4}=\frac{5/8}{4}=\frac{5}{32}.$$

Hence $P(X>\frac{3}{2})=1-F(\frac{3}{2})=\frac{5}{32}$ and $P(\frac{1}{2}\le X\le \frac{3}{2})=F(\frac{3}{2})-F(\frac{1}{2})=\frac{22}{32}=\frac{11}{16}$.

The endpoints do not matter because $F$ is continuous, so by the theorem Point probabilities vanish $P(X=\frac{1}{2})=P(X=\frac{3}{2})=0$; moving an endpoint in or out of the interval adds or removes an event of probability zero.

Sanity check: $f$ is symmetric about $x=1$, so $P(X<\frac{1}{2})$ should equal $P(X>\frac{3}{2})$. Indeed $F(\frac{1}{2})=\frac{5}{32}$, and $\frac{5}{32}+\frac{11}{16}+\frac{5}{32}=1$. ✓

(d) Height against area. $f(1)=\frac{3}{4}\cdot 1\cdot 1=\frac{3}{4}$, while $P(X=1)=0$ by the theorem Point probabilities vanish.

The number $f(1)$ is a rate, probability per unit length near $1$: for a short interval of length $h$ centred at $1$, $P(1-\frac{h}{2}\le X\le 1+\frac{h}{2})\approx \frac{3}{4}h$. The number $P(X=1)$ is the probability of the single point $\{1\}$, which has no length and therefore no area under the curve.

(a) Substitute in the defining integral. By definition $F(-x)={\int }_{-\infty }^{-x}f(t)dt$. Put $t=-s$, so $dt=-ds$; as $t$ runs from $-\infty$ up to $-x$, $s$ runs from $+\infty$ down to $x$. Hence

$$F(-x)={\int }_{\infty }^{x}f(-s)(-ds)={\int }_x^{\infty }f(-s)ds={\int }_x^{\infty }f(s)ds,$$

the last step because $f$ is even. Finally the total area is one, so ${\int }_x^{\infty }f(s)ds=1-{\int }_{-\infty }^{x}f(s)ds=1-F(x)$. $■$

(b) Two specialisations. Put $x=0$ in (a): $F(0)=1-F(0)$, so $F(0)=\frac{1}{2}$, which is exactly the statement that $0$ is a median.

For $a\ge 0$, property (4) of a distribution function gives $P(-a<X\le a)=F(a)-F(-a)$, and the left endpoint may be included because $P(X=-a)=0$ for a continuous distribution function (the theorem Point probabilities vanish). Therefore

$$P(|X|\le a)=P(-a\le X\le a)=F(a)-F(-a)=F(a)-(1-F(a))=2F(a)-1.$$

With $f=\phi$ these are precisely the symmetry identities for the standard normal; the proof shows they have nothing to do with the exponential in $\phi$, only with its evenness.

(c) The Cauchy density. Take $f(x)=\frac{1}{\pi (1+x^2)}$, which is even, positive, and integrates to $\frac{1}{\pi }[\arctan x{]}_{-\infty }^{\infty }=1$. The symmetry argument for $E[X]=0$ says: $xf(x)$ is odd, so its integral over the line is zero. But the definition of $E[X]$ first requires $\int |x|f(x)dx<\infty$, and here

$${\int }_{-\infty }^{\infty }|x|f(x)dx=\frac{2}{\pi }{\int }_0^{\infty }\frac{x}{1+x^2}dx=\frac{1}{\pi }[\ln (1+x^2){]}_0^{\infty }=\infty .$$

So $E[X]$ does not exist at all, and “$0$ by symmetry” is an integral that was never defined. Note that (b) is untouched: the median of the Cauchy distribution is $0$, because a median needs only $F$ and never a moment.

(a) Non-negative with total area one. Clearly $f>0$. By the evenness of $|x|$,

$${\int }_{-\infty }^{\infty }\frac{1}{2}{e}^{-|x|}dx=2{\int }_0^{\infty }\frac{1}{2}{e}^{-x}dx=[-e^{-x}{]}_0^{\infty }=1.$$

(b) Check absolute convergence first. Integrating by parts with $u=x$, $dv=e^{-x}dx$,

$${\int }_{-\infty }^{\infty }|x|f(x)dx={\int }_0^{\infty }x{e}^{-x}dx=[-x{e}^{-x}{]}_0^{\infty }+{\int }_0^{\infty }{e}^{-x}dx=0+1=1<\infty ,$$

so $E[X]$ exists. Now $xf(x)$ is an odd function, so $E[X]=0$ (legitimate precisely because the integral converges absolutely).

For the variance use the computational formula. Since $E[X]=0$, $\mathrm{Var}(X)=E[X^2]$, and by evenness and one more integration by parts ($u=x^2$, $dv=e^{-x}dx$),

$$E[X^2]={\int }_0^{\infty }{x}^2{e}^{-x}dx=[-x^2{e}^{-x}{]}_0^{\infty }+2{\int }_0^{\infty }x{e}^{-x}dx=0+2\cdot 1=2.$$

So $\mathrm{Var}(X)=2$ and $\sigma =\sqrt{2}$.

(c) Integrate from the left in two regimes. For $x<0$ the density is $\frac{1}{2}{e}^t$ on $(-\infty ,x]$, so

$$F(x)={\int }_{-\infty }^x\frac{1}{2}{e}^{t}dt=\frac{1}{2}{e}^x.$$

For $x\ge 0$, the left half contributes $F(0)=\frac{1}{2}$ and the rest is

$$F(x)=\frac{1}{2}+{\int }_0^x\frac{1}{2}{e}^{-t}dt=\frac{1}{2}+\frac{1}{2}\left(1-e^{-x}\right)=1-\frac{1}{2}{e}^{-x}.$$

The lower quartile satisfies $F(x)=\frac{1}{4}<\frac{1}{2}$, so it lies in the $x<0$ branch: $\frac{1}{2}{e}^x=\frac{1}{4}$ gives $x_{1/4}=-\ln 2$. The upper quartile lies in the $x\ge 0$ branch: $1-\frac{1}{2}{e}^{-x}=\frac{3}{4}$ gives $e^{-x}=\frac{1}{2}$, so $x_{3/4}=\ln 2$. They are mirror images, as they must be for an even density.

(d) Law of the unconscious statistician. Split at $0$ so that $|x|$ can be removed:

$$E\left[e^{X/2}\right]=\frac{1}{2}{\int }_{-\infty }^0{e}^{x/2}{e}^{x}dx+\frac{1}{2}{\int }_0^{\infty }{e}^{x/2}{e}^{-x}dx=\frac{1}{2}{\int }_{-\infty }^0{e}^{3x/2}dx+\frac{1}{2}{\int }_0^{\infty }{e}^{-x/2}dx=\frac{1}{2}\cdot \frac{2}{3}+\frac{1}{2}\cdot 2=\frac{4}{3}.$$

For $E[e^X]$ the right-hand integrand is $\frac{1}{2}{e}^x{e}^{-x}=\frac{1}{2}$, a constant, and ${\int }_0^{\infty }\frac{1}{2}dx$ diverges. The expectation is $+\infty$: the density's tail $e^{-x}$ decays exactly as fast as $e^x$ grows, so nothing is left to make the integral converge.

(a) Constant density, linear cdf. The interval has length $4$, so $f(x)=\frac{1}{4}$ for $-1\le x\le 3$ and $0$ otherwise. Then $F(x)=0$ for $x<-1$, $F(x)=\frac{x+1}{4}$ for $-1\le x\le 3$, and $F(x)=1$ for $x>3$. By the uniform formulas,

$$E[X]=\frac{-1+3}{2}=1,\mathrm{Var}(X)=\frac{(3-(-1){)}^2}{12}=\frac{16}{12}=\frac{4}{3}.$$

(b) Turn each event into an interval. $X^2<1$ means $-1<X<1$, so

$$P(X^2<1)=F(1)-F(-1)=\frac{2}{4}-0=\frac{1}{2}.$$

$|X-1|>1$ means $X<0$ or $X>2$, two disjoint events:

$$P(|X-1|>1)=F(0)+(1-F(2))=\frac{1}{4}+\left(1-\frac{3}{4}\right)=\frac{1}{2}.$$

Both answers are also visible as lengths: each event occupies half of the width-$4$ interval.

(c) Invert the cdf. Solving $F(x)=u$, that is $\frac{x+1}{4}=u$, gives $x=4u-1$, so by the inverse transform theorem $X=4U-1$ has distribution function $F$. Check: $E[4U-1]=4\cdot \frac{1}{2}-1=1$ and $\mathrm{Var}(4U-1)=16\cdot \frac{1}{12}=\frac{4}{3}$, agreeing with (a).

(d) A non-linear inverse. The density $3w^2$ on $[0,1]$ has cdf $F_W(w)={\int }_0^{w}3t^{2}dt=w^3$, which is continuous and strictly increasing on $[0,1]$, so the theorem applies. Solving $w^3=u$ gives $F_W^{-1}(u)=u^{1/3}$, hence

$$W=U^{1/3}.$$

From the density: $P(W\le \frac{1}{2})=F_W(\frac{1}{2})=\frac{1}{8}$. From $U$: $P(U^{1/3}\le \frac{1}{2})=P(U\le \frac{1}{8})=\frac{1}{8}$, since $P(U\le u)=u$. They agree. Note the cube root pushes uniform draws upward — only one draw in eight lands below $\frac{1}{2}$ — which is what a density that grows like $3w^2$ towards $w=1$ requires.

(a) Differentiate the exponential. Write $f(x)=C\exp \left(-\frac{(x-\mu {)}^2}{2{\sigma }^2}\right)$ with $C=\frac{1}{\sigma \sqrt{2\pi }}$. By the chain rule,

$$f^{\prime }(x)=-\frac{x-\mu }{{\sigma }^2}f(x).$$

Since $f(x)>0$ everywhere, $f^{\prime }$ has the sign of $-(x-\mu )$: positive for $x<\mu$, zero at $x=\mu$, negative for $x>\mu$. So $f$ rises to a single maximum at $x=\mu$, where the exponential is $e^0=1$ and the height is $C=\frac{1}{\sigma \sqrt{2\pi }}$.

Differentiating again with the product rule,

$$f^{\prime \prime }(x)=-\frac{1}{{\sigma }^2}f(x)-\frac{x-\mu }{{\sigma }^2}{f}^{\prime }(x)=-\frac{1}{{\sigma }^2}f(x)+\frac{(x-\mu {)}^2}{{\sigma }^4}f(x)=\frac{(x-\mu {)}^2-{\sigma }^2}{{\sigma }^4}f(x).$$

This vanishes exactly at $x=\mu \pm \sigma$, is negative between them (concave, the cap of the bell) and positive outside (convex, the tails). The sign changes, so these are genuine inflection points, at height $f(\mu \pm \sigma )=\frac{e^{-1/2}}{\sigma \sqrt{2\pi }}$.

(b) Set the densities equal. With $\mu =0$,

$$\frac{1}{\sqrt{2\pi }}{e}^{-x^2/2}=\frac{1}{2\sqrt{2\pi }}{e}^{-x^2/8}⟺e^{-x^2/2+x^2/8}=\frac{1}{2}⟺e^{-3x^2/8}=\frac{1}{2}.$$

Taking logarithms, $\frac{3x^2}{8}=\ln 2$, so the crossings are at

$$x=\pm \sqrt{\frac{8\ln 2}{3}}\approx \pm \mathrm{1.36.}$$

At $x=0$ the narrow curve is higher ($f_1(0)\approx 0.399$ against $f_2(0)\approx 0.199$), so $f_1$ is above $f_2$ between the crossings and below it outside them: the narrow bell wins in the middle and the wide bell wins in both tails.

(c) The drawing. Both curves are symmetric about $x=0$, where each has its peak: $f_1(0)=\frac{1}{\sqrt{2\pi }}\approx 0.399$ and $f_2(0)=\frac{1}{2\sqrt{2\pi }}\approx 0.199$, exactly half. By (a), $f_1$ has inflection points at $x=\pm 1$ (height $\approx 0.242$) and $f_2$ at $x=\pm 2$ (height $\approx 0.121$). Mark the crossings at $x\approx \pm 1.36$, where both curves have height $\approx 0.158$. The tails: at $x=\pm 3$, $f_1\approx 0.004$ is already invisible while $f_2\approx 0.065$ is still clearly off the axis, and at $x=\pm 5$ both are negligible.

Shaded areas. Under $f_1$ the region over $[-1,1]$ has area $P(|Z|\le 1)=2\Phi (1)-1\approx 0.683$. Under $f_2$, the variable is $X=2Z$, so the same region has area $P(|2Z|\le 1)=P(|Z|\le \frac{1}{2})=2\Phi (\frac{1}{2})-1\approx 0.383$. The narrow curve's shaded area is larger, as the picture shows: it is taller over the whole of $[-1,1]$.

Yet both total areas are $1$. The wide curve is half as tall but twice as wide: $f_2(x)=\frac{1}{2}{f}_1(x/2)$, so substituting $x=2t$ gives $\int f_2(x)dx=\int \frac{1}{2}{f}_1(t)\cdot 2dt=\int f_1(t)dt=1$. What $f_2$ loses over $[-1,1]$ it recovers in the tails beyond the crossing points.

(a) Standardise, then read left tails. For $X$: $z=\frac{53-50}{3}=1$, so

$$P(X>53)=1-\Phi (1)=1-0.8413=\mathrm{0.1587.}$$

For $Y$ the endpoints standardise to $z=\frac{38-44}{4}=-1.5$ and $z=\frac{50-44}{4}=1.5$, a symmetric band, so by the symmetry identity

$$P(38\le Y\le 50)=2\Phi (1.5)-1=2(0.9332)-1=\mathrm{0.8664.}$$

Sanity check: a $\pm 1.5\sigma$ band should hold more than the $\pm 1\sigma$ band ($0.683$) and less than the $\pm 2\sigma$ band ($0.954$). ✓

(b) The inverse problem. We need $z$ with $\Phi (z)=0.1056$. That is below $\frac{1}{2}$, so $z<0$, and $\Phi (-z)=1-0.1056=0.8944=\Phi (1.25)$, giving $z=-1.25$. Untransform with the standard deviation, not the variance:

$$w=\mu +z\sigma =50-1.25\times 3=\mathrm{46.25.}$$

(c) The difference of two independent normals. Let $D=X-Y$. By the theorem on linear combinations of independent normals, with $a=1$, $b=-1$,

$$D\sim N\left(50-44,3^2+4^2\right)=N(6,25),{\sigma }_D=5.$$

The variances add even though the variables are subtracted. The event “$A$ lighter than $B$” is $D<0$, so

$$P(D<0)=\Phi \left(\frac{0-6}{5}\right)=\Phi (-1.2)=1-\Phi (1.2)=1-0.8849=\mathrm{0.1151.}$$

(d) Sums and averages. By the corollary on sums of an independent sample, $T=Y_1+\cdots +Y_4\sim N(4\times 44,4\times 16)=N(176,64)$, so ${\sigma }_T=8$ — not $4\times 4=16$. Then

$$P(T>188)=1-\Phi \left(\frac{188-176}{8}\right)=1-\Phi (1.5)=\mathrm{0.0668.}$$

The average of $n$ bags has standard deviation $\sigma /\sqrt{n}=4/\sqrt{n}$; setting this equal to $1$ gives $\sqrt{n}=4$, so $n=16$. Quadrupling the sample halves the spread, and the spread here must be quartered.

(a) The rate is $3$, so the mean is $\frac{1}{3}$. By the theorem Poisson counts and exponential gaps, $X\sim \text{Exp}(3)$:

$$f(x)=3e^{-3x}(x\ge 0),P(X>x)=e^{-3x}.$$

The mean is $1/\lambda =\frac{1}{3}$ hour $=20$ minutes, and the standard deviation equals the mean for an exponential, so it is also $20$ minutes. Finally $P(X>\frac{1}{3})=e^{-3\cdot \frac{1}{3}}=e^{-1}$: the chance of waiting longer than the average wait is about $0.37$, well under a half.

(b) The median. Solve $1-e^{-3m}=\frac{1}{2}$: $e^{-3m}=\frac{1}{2}$, so $m=\frac{\ln 2}{3}$ hours $=20\ln 2$ minutes, a little under $14$ minutes.

It lies below the mean because (a) showed $P(X>\text{mean})=e^{-1}<\frac{1}{2}$: more than half of the probability sits below the mean, so the point that splits the probability in half must be to its left. This is the signature of a right-skewed distribution — most waits are short, and the long ones drag the average up without moving the median.

(c) Memorylessness. In hours, the question is $P(X>\frac{1}{2}\mid X>\frac{1}{6})$. By the theorem Memorylessness with $s=\frac{1}{6}$ and $t=\frac{1}{3}$,

$$P\left(X>\frac{1}{6}+\frac{1}{3}|X>\frac{1}{6}\right)=P\left(X>\frac{1}{3}\right)=e^{-1}.$$

Ten minutes of waiting has bought nothing: the remaining wait is distributed exactly as it was on arrival.

(d) Two more survival calculations. With $m=\ln 2/3$ and mean $\frac{1}{3}$,

$$P\left(m<X<\frac{1}{3}\right)=P(X>m)-P\left(X>\frac{1}{3}\right)=e^{-\ln 2}-e^{-1}=\frac{1}{2}-e^{-1}\approx \mathrm{0.132.}$$

For the $40$-minute window, $t=\frac{2}{3}$ hour and the count $N(\frac{2}{3})$ is Poisson with mean $\lambda t=2$, so $P(N(\frac{2}{3})=0)=e^{-2}$. The survival function gives $P(X>\frac{2}{3})=e^{-3\cdot \frac{2}{3}}=e^{-2}$, the same number — as it must be, since “no bus by time $t$” and “the wait exceeds $t$” are the same event.

Why survival functions suffice. $P(X>t)=1-F(t)$, so the survival function determines the distribution function and hence, by the theorem that density and distribution determine each other, the whole distribution. Showing that a variable has survival function $e^{-\nu t}$ for $t\ge 0$ therefore proves it is $\text{Exp}(\nu )$.

(a) Rescaling. For $t\ge 0$ and $c>0$,

$$P(cX>t)=P\left(X>\frac{t}{c}\right)=e^{-\lambda t/c}=e^{-(\lambda /c)t},$$

which is the survival function of $\text{Exp}(\lambda /c)$. $■$ As a check on the direction, $E[cX]=c/\lambda =1/(\lambda /c)$, consistent with the rule that the mean is the reciprocal of the rate: stretching the time scale by $c$ divides the rate by $c$.

(b) The minimum of independent exponentials. The minimum exceeds $t$ exactly when both variables do, so $\{M>t\}=\{X>t\}\cap \{Y>t\}$. By independence the probabilities multiply:

$$P(M>t)=P(X>t)P(Y>t)=e^{-\lambda t}{e}^{-\mu t}=e^{-(\lambda +\mu )t},t\ge 0.$$

This is the survival function of $\text{Exp}(\lambda +\mu )$, so $M\sim \text{Exp}(\lambda +\mu )$ and $E[M]=\frac{1}{\lambda +\mu }$. $■$ Note that this is smaller than both $1/\lambda$ and $1/\mu$: waiting for the first of two things is faster than waiting for either alone, and the rates simply add.

(c) Constant hazard. With $f(t)=\lambda e^{-\lambda t}$ and $P(X>t)=e^{-\lambda t}$,

$$h(t)=\frac{\lambda e^{-\lambda t}}{e^{-\lambda t}}=\lambda \text{for every }t\ge 0.$$

The hazard is the instantaneous failure rate given survival so far: for small $\delta >0$,

$$P(t<X\le t+\delta \mid X>t)=\frac{F(t+\delta )-F(t)}{P(X>t)}\approx \frac{f(t)\delta }{P(X>t)}=h(t)\delta .$$

A constant hazard says the chance of failing in the next instant does not depend on the age $t$ already reached, which is exactly what memorylessness asserts: indeed the memoryless identity gives $P(t<X\le t+\delta \mid X>t)=1-P(X>\delta )=1-e^{-\lambda \delta }\approx \lambda \delta$ directly. Ageing would show up as a hazard that increases with $t$.

(d) Apply (b). The device's lifetime is $\min (X,Y)$ with $\lambda =1$ and $\mu =2$, so it is $\text{Exp}(3)$. Then

$$P(\text{working after }1000\text{ h})=P(M>1)=e^{-3}\approx 0.050,$$

and the expected lifetime is $\frac{1}{1+2}=\frac{1}{3}$ thousand hours, about $333$ hours — shorter than the $500$ hours of the weaker component on its own.

(a) A chi-squared with three degrees of freedom. By definition $V\sim {\chi }_3^2$, so $E[V]=3$ and $\mathrm{Var}(V)=2\cdot 3=6$. From the computational formula for the variance,

$$E[V^2]=\mathrm{Var}(V)+(E[V]{)}^2=6+9=15.$$

Expanding instead, $V^2={\sum }_i{Z}_i^4+2{\sum }_{i<j}{Z}_i^2{Z}_j^2$, with three fourth-power terms and three cross terms. Independence gives $E[Z_i^2{Z}_j^2]=E[Z_i^2]E[Z_j^2]=1$ for $i\ne j$, so

$$E[V^2]=3\cdot 3+2\cdot 3\cdot 1=9+6=15.$$

The two routes agree; the cross terms are where independence enters.

(b) Fold the normal. For $w\ge 0$, by the symmetry identities,

$$F_W(w)=P(-w\le Z\le w)=2\Phi (w)-1,f_W(w)=F_W^{\prime }(w)=2\phi (w)=\sqrt{\frac{2}{\pi }}{e}^{-w^2/2},$$

and $F_W(w)=0$ for $w<0$. Both halves of the bell are folded onto $[0,\infty )$, which is why the density doubles. For the mean, the substitution is immediate:

$$E[W]={\int }_0^{\infty }w\cdot \frac{2}{\sqrt{2\pi }}{e}^{-w^2/2}dw=\frac{2}{\sqrt{2\pi }}[-e^{-w^2/2}{]}_0^{\infty }=\frac{2}{\sqrt{2\pi }}=\sqrt{\frac{2}{\pi }}.$$

Since $W^2=Z^2$, $E[W^2]=E[Z^2]=1$, and so

$$\mathrm{Var}(W)=1-\frac{2}{\pi }\approx \mathrm{0.363.}$$

(c) The square of a standard normal. $P(Z^2\le 1)=P(|Z|\le 1)=2\Phi (1)-1$ (about $0.683$). For the median $m$ we need $P(Z^2\le m)=2\Phi (\sqrt{m})-1=\frac{1}{2}$, so $\Phi (\sqrt{m})=\frac{3}{4}$ and

$$m={\left({\Phi }^{-1}\left(\frac{3}{4}\right)\right)}^2\approx \mathrm{0.455.}$$

This is below $E[V_1]=1$: the ${\chi }_1^2$ density is unbounded at $0$ and has a long right tail, so it is strongly right-skewed and the median sits well below the mean.

(d) The denominator of a $t$. By linearity and the scaling of variance, with $E[V]=\nu$ and $\mathrm{Var}(V)=2\nu$,

$$E\left[\frac{V}{\nu }\right]=1,\mathrm{Var}\left(\frac{V}{\nu }\right)=\frac{2\nu }{{\nu }^2}=\frac{2}{\nu }⟶0.$$

So $V/\nu$ has mean $1$ and a standard deviation $\sqrt{2/\nu }$ that shrinks to zero: it concentrates at $1$. In $T=Z/\sqrt{V/\nu }$ the denominator therefore tends to $1$ and $T$ behaves like the numerator $Z$ alone, which is $N(0,1)$. For finite $\nu$ the denominator still wobbles, and the wobble is what fattens the tails of $t_{\nu }$.

(a) Both halves of the validity condition. $np=25\times \frac{1}{2}=12.5\ge 10$ and $n(1-p)=12.5\ge 10$, so the rule is satisfied. The approximating normal has

$$\mu =np=12.5,\sigma =\sqrt{np(1-p)}=\sqrt{25\cdot \frac{1}{2}\cdot \frac{1}{2}}=\sqrt{\frac{25}{4}}=\frac{5}{2}.$$

(b) Where the bar for $15$ ends. The bars up to and including $k=15$ end at $15.5$, so with the correction

$$P(X\le 15)\approx \Phi \left(\frac{15.5-12.5}{2.5}\right)=\Phi (1.2)=\mathrm{0.8849.}$$

Cutting at $15$ instead gives $z=\frac{15-12.5}{2.5}=1$ and $\Phi (1)=0.8413$.

Against the exact $0.8852$, the corrected value is off by $0.0003$ and the uncorrected one by $0.044$ — more than a hundred times worse. The uncorrected version has thrown away half of the bar at $15$, whose area is about $0.097$, which accounts for almost the whole of its error. Sanity check: $15$ is above the mean, so the answer must exceed $\frac{1}{2}$. ✓

(c) Two whole bars. The event $\{12\le X\le 13\}$ is the pair of bars spanning $[11.5,13.5]$, so

$$P(12\le X\le 13)\approx \Phi \left(\frac{13.5-12.5}{2.5}\right)-\Phi \left(\frac{11.5-12.5}{2.5}\right)=\Phi (0.4)-\Phi (-0.4)=2\Phi (0.4)-1=0.3108,$$

against the exact $0.3100$. Without the correction the “interval” $[12,13]$ would have width one and would cover only half of each bar. Note that the mean $12.5$ sits exactly on the boundary between the two bars, so by symmetry $P(X=12)=P(X=13)$ and each is about $0.155$.

(d) The inverse problem. With $X\sim \text{Binomial}(n,\frac{1}{2})$, $\mu =\frac{n}{2}$ and $\sigma =\sqrt{n/4}=\frac{\sqrt{n}}{2}$. The event $\left|\frac{X}{n}-\frac{1}{2}\right|\le 0.05$ is $|X-\frac{n}{2}|\le 0.05n$, and standardising,

$$P\left(\left|\frac{X}{n}-\frac{1}{2}\right|\le 0.05\right)\approx P\left(|Z|\le \frac{0.05n}{\sqrt{n}/2}\right)=P\left(|Z|\le 0.1\sqrt{n}\right)=2\Phi \left(0.1\sqrt{n}\right)-1.$$

We need this to be at least $0.9544=2(0.9772)-1=2\Phi (2)-1$, that is $\Phi (0.1\sqrt{n})\ge \Phi (2)$. Since $\Phi$ is strictly increasing this holds exactly when $0.1\sqrt{n}\ge 2$, i.e. $\sqrt{n}\ge 20$, so the smallest $n$ is

$$n=400.$$

Including the continuity correction would only enlarge the band (to $|X-\frac{n}{2}|\le 0.05n+\frac{1}{2}$) and so only raise the probability, so $n=400$ certainly suffices under either convention. The $\sqrt{n}$ is the whole story: to halve the tolerance to $0.025$ at the same confidence would need $n=1600$, four times as many tosses.

(a) Argument A ignores the independence hypothesis. The definition of the $t$-distribution requires the numerator $Z$ and the chi-squared $V$ to be independent. Here $V=Z_1^2+Z_2^2$ is a function of $Z_1$, so the pair is not independent and the definition simply does not apply. The conclusion is also false, and the cleanest way to see it is that $T$ is bounded:

$$T^2=\frac{Z_1^2}{(Z_1^2+Z_2^2)/2}=\frac{2Z_1^2}{Z_1^2+Z_2^2}\le 2,$$

because the numerator is at most the denominator. So $|T|\le \sqrt{2}$ with probability one.

No $t$ variable is bounded. Let $T^{\prime }=Z/\sqrt{V/\nu }$ with $Z$ and $V\sim {\chi }_{\nu }^2$ independent, and fix any $c>0$. On the event $\{Z>2c\}\cap \{V<4\nu \}$ we have $\sqrt{V/\nu }<2$ and hence $T^{\prime }>2c/2=c$. The two events are independent, $P(Z>2c)=1-\Phi (2c)>0$ since $\Phi <1$ everywhere, and $P(V<4\nu )>0$ since $V\ge 0$ has mean $\nu$ and so cannot exceed $4\nu$ with probability one. Their intersection therefore has positive probability, and $P(T^{\prime }>c)>0$ for every $c$: $T^{\prime }$ exceeds any bound. A variable confined to $[-\sqrt{2},\sqrt{2}]$ cannot be $t_2$.

The repair is to use a chi-squared built from other normals: with $Z_3,Z_4$ independent of $Z_1$, the variable $Z_1/\sqrt{(Z_3^2+Z_4^2)/2}$ is genuinely $t_2$, because now the numerator and denominator are independent.

(b) Argument B swaps expectation and squaring. The step $E[(Z^2{)}^2]=(E[Z^2]{)}^2$ is false: expectation does not commute with a non-linear function, and $E[g(Y)]=g(E[Y])$ holds only for linear $g$. In fact $E[Y^2]-(E[Y]{)}^2=\mathrm{Var}(Y)\ge 0$ with equality only for a constant $Y$, and $Y=Z^2$ is plainly not constant — so the “conclusion” is the very thing that shows the step was wrong.

Correctly, integrate by parts with $u=z^3$ and $dv=z{e}^{-z^2/2}dz$, so $v=-e^{-z^2/2}$:

$$E[Z^4]=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }{z}^3\cdot z{e}^{-z^2/2}dz=\frac{1}{\sqrt{2\pi }}\left([-z^3{e}^{-z^2/2}{]}_{-\infty }^{\infty }+3{\int }_{-\infty }^{\infty }{z}^2{e}^{-z^2/2}dz\right)=0+3E[Z^2]=3.$$

Hence $\mathrm{Var}(Z^2)=3-1=2$, which is the case $k=1$ of $\mathrm{Var}({\chi }_k^2)=2k$.

(c) Argument C ignores independence in the definition of ${\chi }^2$. The claim $W\sim N(0,1)$ is correct: $P(-Z\le z)=P(Z\ge -z)=1-\Phi (-z)=\Phi (z)$ by the symmetry identity. The false step is “hence ${\chi }_2^2$”: the definition requires the standard normals being squared to be independent, and $W=-Z$ is a function of $Z$.

In fact $Z^2+W^2=Z^2+Z^2=2Z^2$. By linearity, $E[2Z^2]=2$, so the quoted mean is right — the mean never needed independence. But by the scaling of variance and (b),

$$\mathrm{Var}(2Z^2)=4\mathrm{Var}(Z^2)=4\cdot 2=8,$$

not $4$. Of the three quoted facts — the distribution, the mean and the variance — only the mean survives. (The distribution is wrong too: $P(2Z^2\le 2)=P(|Z|\le 1)=2\Phi (1)-1$, and there is no reason for a ${\chi }_2^2$ variable to match that.)

(a) Exchange the order of integration. Since $X\ge 0$, the density vanishes on $(-\infty ,0)$ and

$$E[X]={\int }_0^{\infty }xf(x)dx={\int }_0^{\infty }\left({\int }_0^{x}1dt\right)f(x)dx={\iint }_{\{0\le t\le x\}}f(x)dtdx.$$