Exam P — Transformations & Order Statistics Flashcards

Write the cdf of $Y$ directly in terms of $X$: $F_Y(y)=\Pr(Y\leq y)=\Pr(g(X)\leq y)$. Solve the event $\{g(X)\leq y\}$ for $X$ to express it through $F_X$, then differentiate to get the pdf: $f_Y(y)=\frac{d}{dy}F_Y(y)$. It works for any $g$ (monotone or not) and is the safest general approach.

For $0<y<4$: $F_Y(y)=\Pr(X^{2}\leq y)=\Pr(X\leq\sqrt{y})=\frac{\sqrt{y}}{2}$ since $F_X(x)=\frac{x}{2}$. Differentiate: $f_Y(y)=\frac{d}{dy}\frac{\sqrt{y}}{2}=\frac{1}{4\sqrt{y}}$ for $0<y<4$ (and $0$ otherwise).

If $g$ is monotone with inverse $x=g^{-1}(y)$, then $f_Y(y)=f_X\!\big(g^{-1}(y)\big)\left|\frac{dx}{dy}\right|$. The factor $\left|\frac{dx}{dy}\right|$ is the (one-dimensional) Jacobian; the absolute value keeps the density nonnegative. Forgetting this factor is the most common transformation error.

Invert: $x=y^{1/3}$, so $\frac{dx}{dy}=\frac{1}{3}y^{-2/3}$. Since $f_X(x)=1$ on $(0,1)$, $f_Y(y)=1\cdot\left|\tfrac{1}{3}y^{-2/3}\right|=\frac{1}{3}y^{-2/3}$ for $0<y<1$. Check: $\int_{0}^{1}\frac{1}{3}y^{-2/3}\,dy=\big[y^{1/3}\big]_{0}^{1}=1$.

Here $Y\in(0,1)$ and $x=-\ln y$, so $\left|\frac{dx}{dy}\right|=\frac{1}{y}$. Then $f_Y(y)=f_X(-\ln y)\cdot\frac{1}{y}=e^{\ln y}\cdot\frac{1}{y}=y\cdot\frac{1}{y}=1$ for $0<y<1$. So $Y\sim U(0,1)$ — an instance of the probability integral transform since $e^{-X}=1-F_X(X)$.

If $X$ is continuous with cdf $F_X$, then $U=F_X(X)\sim U(0,1)$. Conversely, if $U\sim U(0,1)$ then $X=F_X^{-1}(U)$ has cdf $F_X$ — this is the basis of inverse-transform simulation.

$g(x)=x^{2}$ is not monotone on $(-1,1)$: both $x=\pm\sqrt{y}$ map to the same $y$, so you must **sum over both branches**. $f_Y(y)=f_X(\sqrt{y})\cdot\frac{1}{2\sqrt{y}}+f_X(-\sqrt{y})\cdot\frac{1}{2\sqrt{y}}=\frac{1}{2}\cdot\frac{1}{2\sqrt{y}}+\frac{1}{2}\cdot\frac{1}{2\sqrt{y}}=\frac{1}{2\sqrt{y}}$ for $0<y<1$.

Scaling an exponential by a positive constant gives another exponential: $Y=2X\sim\text{Exp}$ with mean $2\cdot 5=10$. Then $\Pr(Y>15)=e^{-15/10}=e^{-1.5}\approx 0.2231$.

For $y>0$: $F_Y(y)=\Pr(-2\ln U\leq y)=\Pr\!\big(\ln U\geq -\tfrac{y}{2}\big)=\Pr\!\big(U\geq e^{-y/2}\big)=1-e^{-y/2}$. That is the cdf of $\text{Exp}$ with mean $2$ (rate $\tfrac{1}{2}$). $\Pr(Y>4)=e^{-4/2}=e^{-2}\approx 0.1353$.

Invert: $z=\frac{x-\mu}{\sigma}$, $\left|\frac{dz}{dx}\right|=\frac{1}{\sigma}$. $f_X(x)=\frac{1}{\sqrt{2\pi}}e^{-z^{2}/2}\cdot\frac{1}{\sigma}=\frac{1}{\sigma\sqrt{2\pi}}\exp\!\Big(-\frac{(x-\mu)^{2}}{2\sigma^{2}}\Big)$, which is the $N(\mu,\sigma^{2})$ density.

With inverse $x=x(u,v)$, $y=y(u,v)$, the joint density transforms as $f_{U,V}(u,v)=f_{X,Y}\big(x(u,v),y(u,v)\big)\,|J|$, where the Jacobian determinant is $J=\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}$. Always take $|J|$ and re-express the support in the new variables.

Inverse: $x=v$, $y=u-v$; the Jacobian is $|J|=1$. $f_{U,V}(u,v)=e^{-v}e^{-(u-v)}\cdot 1=e^{-u}$ on the region $0<v<u$. Marginalize over $v$: $f_U(u)=\int_{0}^{u}e^{-u}\,dv=u\,e^{-u}$ for $u>0$ — the $\text{Gamma}(2,1)$ density.

$f_Z(z)=\int_{-\infty}^{\infty} f_X(x)\,f_Y(z-x)\,dx=\int_{-\infty}^{\infty} f_X(z-y)\,f_Y(y)\,dy$. Care is needed with limits: integrate only where **both** densities are positive.

$f_Z(z)=\int f_X(x)f_Y(z-x)\,dx$ with both factors equal to $1$ when $0<x<1$ and $0<z-x<1$. For $0\leq z\leq 1$: limits $0<x<z$ give $f_Z(z)=z$. For $1<z\leq 2$: limits $z-1<x<1$ give $f_Z(z)=2-z$. This is the triangular density on $(0,2)$ peaking at $z=1$.

The event $\{X+Y\leq 0.5\}$ is the triangular region in the unit square below the line $x+y=0.5$. Area $=\frac{1}{2}(0.5)(0.5)=0.125$. Since the joint density is $1$ on the square, $\Pr(X+Y\leq 0.5)=0.125$.

The region $\{x+y>1.5\}$ inside the unit square is the small triangle with vertices $(0.5,1),(1,0.5),(1,1)$, with legs of length $0.5$. Area $=\frac{1}{2}(0.5)(0.5)=0.125$, so $\Pr(X+Y>1.5)=0.125$.

Discrete convolution: $\Pr(S=s)=\sum_{k}\Pr(X=k)\,\Pr(Y=s-k)$, summing over all $k$ with both terms positive. For two fair dice each have $\Pr=\tfrac{1}{6}$ on $\{1,\dots,6\}$. For $s=5$ the valid pairs are $k\in\{1,2,3,4\}$, so $\Pr(S=5)=\sum_{k=1}^{4}\tfrac{1}{6}\cdot\tfrac{1}{6}=\frac{4}{36}=\frac{1}{9}$. The sum's pmf is the triangular pattern $\tfrac{1}{36},\tfrac{2}{36},\dots,\tfrac{6}{36},\dots,\tfrac{1}{36}$ for $s=2,\dots,12$ — the discrete analogue of the continuous triangular density.

$\Pr(S=s)=\sum_{k=0}^{s}\Pr(X=k)\Pr(Y=s-k)=\sum_{k=0}^{s}\frac{e^{-\lambda_1}\lambda_1^{k}}{k!}\cdot\frac{e^{-\lambda_2}\lambda_2^{\,s-k}}{(s-k)!}$. Factor out $e^{-(\lambda_1+\lambda_2)}$ and multiply/divide by $s!$: $\Pr(S=s)=\frac{e^{-(\lambda_1+\lambda_2)}}{s!}\sum_{k=0}^{s}\binom{s}{k}\lambda_1^{k}\lambda_2^{\,s-k}=\frac{e^{-(\lambda_1+\lambda_2)}(\lambda_1+\lambda_2)^{s}}{s!}$ by the binomial theorem. That is the $\text{Poisson}(\lambda_1+\lambda_2)$ pmf, so Poisson rates add under independent sums.

The sum is $\text{Gamma}(\alpha=n,\lambda)$, also called Erlang-$n$. Reason: each exponential has MGF $\frac{\lambda}{\lambda-t}$, and independent MGFs multiply, giving $\big(\frac{\lambda}{\lambda-t}\big)^{n}$ — the $\text{Gamma}(n,\lambda)$ MGF. Mean $=\frac{n}{\lambda}$, variance $=\frac{n}{\lambda^{2}}$.

Each delay has rate $\lambda=3$, so $S\sim\text{Gamma}(10,3)$. Mean $=\frac{n}{\lambda}=\frac{10}{3}\approx 3.33$. Variance $=\frac{n}{\lambda^{2}}=\frac{10}{9}\approx 1.11$.

$\frac{X}{Y}<1\iff X<Y$. By symmetry of the unit square about the line $x=y$, $\Pr(X<Y)=\frac{1}{2}$ (the diagonal contributes zero probability).

$Y\geq 0$; invert $x=1-e^{-y}$, $\frac{dx}{dy}=e^{-y}$. $f_Y(y)=f_X(1-e^{-y})\cdot e^{-y}=1\cdot e^{-y}=e^{-y}$, $y>0$. So $Y\sim\text{Exp}(1)$. (Again the inverse-cdf transform, since $1-e^{-y}$ is the $\text{Exp}(1)$ cdf.)

$F_{(n)}(x)=\Pr(\text{all}\leq x)=[F(x)]^{n}$. Differentiate: $f_{(n)}(x)=n[F(x)]^{n-1}f(x)$.

$\Pr(X_{(1)}>x)=\Pr(\text{all}>x)=[1-F(x)]^{n}$, so $F_{(1)}(x)=1-[1-F(x)]^{n}$. Differentiate: $f_{(1)}(x)=n[1-F(x)]^{n-1}f(x)$.

$f_{(k)}(x)=\frac{n!}{(k-1)!\,(n-k)!}\,[F(x)]^{k-1}\,[1-F(x)]^{n-k}\,f(x)$. Interpretation: $k-1$ values fall below $x$, one is at $x$, and $n-k$ exceed $x$. Dropping the multinomial coefficient is a classic error.

$\Pr(\min>x)=\prod_i e^{-\lambda_i x}=e^{-(\sum_i\lambda_i)x}$, so $\min_i X_i\sim\text{Exp}\!\big(\sum_i\lambda_i\big)$. The rate is the **sum** of the individual rates (not the average). For iid $\text{Exp}(\lambda)$, the minimum is $\text{Exp}(n\lambda)$.

Rates: $\lambda_1=\tfrac{1}{4},\ \lambda_2=\tfrac{1}{5},\ \lambda_3=\tfrac{1}{10}$. First failure $=\min\sim\text{Exp}(\lambda_1+\lambda_2+\lambda_3)$, rate $=0.25+0.20+0.10=0.55$. Expected time $=\frac{1}{0.55}\approx 1.818$ years.

$\Pr(X_1<\text{all others})=\dfrac{\lambda_1}{\lambda_1+\lambda_2+\cdots+\lambda_n}$. Example: if $\lambda_1=2$ and $\lambda_2=3$, then $\Pr(X_1\text{ first})=\frac{2}{2+3}=0.4$.

The minimum of five iid $\text{Exp}(0.2)$ is $\text{Exp}(5\cdot 0.2)=\text{Exp}(1)$. $\Pr(\min>3)=e^{-1\cdot 3}=e^{-3}\approx 0.0498$.

Since $F(t)=t$: $\Pr(X_{(n)}<t)=t^{n}$, and $E[X_{(n)}]=\frac{n}{n+1}$. For $n=3$, $t=0.7$: $\Pr(X_{(3)}<0.7)=0.7^{3}=0.343$.

For $U(0,1)$, $E[X_{(k)}]=\frac{k}{n+1}$. Minimum: $E[X_{(1)}]=\frac{1}{5}=0.2$. Maximum: $E[X_{(4)}]=\frac{4}{5}=0.8$.

Substituting $F(x)=x$, $f(x)=1$ into the $k$-th order-statistic pdf gives $f_{(k)}(x)=\frac{n!}{(k-1)!(n-k)!}x^{k-1}(1-x)^{n-k}$ — the $\text{Beta}(k,n-k+1)$ density. Hence $E[X_{(k)}]=\frac{k}{n+1}$ and $\operatorname{Var}(X_{(k)})=\frac{k(n-k+1)}{(n+1)^{2}(n+2)}$.

Here $n=5$, $k=3$, so $X_{(3)}\sim\text{Beta}(3,3)$. Mean $=\frac{k}{n+1}=\frac{3}{6}=0.5$. Variance $=\frac{k(n-k+1)}{(n+1)^{2}(n+2)}=\frac{3\cdot 3}{6^{2}\cdot 7}=\frac{9}{252}\approx 0.0357$.

System failure $=\max(X_1,X_2)$ of two iid $\text{Exp}(\lambda)$, $\lambda=\tfrac{1}{10}$. Using $E[\max]=\frac{1}{\lambda}\big(1+\tfrac{1}{2}\big)=\frac{3}{2\lambda}$ (mean is sum of independent $\text{Exp}$ gaps): $E[\max]=10\cdot 1.5=15$ hours.