Exam P — Multivariate Distributions Flashcards

Integrate the joint pdf over the region: $\Pr((X,Y)\in R)=\iint_{R} f(x,y)\,dx\,dy$. A valid joint pdf satisfies $f(x,y)\geq 0$ and $\iint f(x,y)\,dx\,dy=1$ over the whole support.

Require the density to integrate to $1$: $\int_{0}^{1}\int_{0}^{1} c(x+y)\,dx\,dy=1$. Inner: $\int_{0}^{1}(x+y)\,dx=\tfrac{1}{2}+y$. Outer: $\int_{0}^{1}\left(\tfrac{1}{2}+y\right)dy=\tfrac{1}{2}+\tfrac{1}{2}=1$. So $c\cdot 1=1\Rightarrow c=1$.

Integrate over the triangle ($x$ runs $0$ to $y$, then $y$ runs $0$ to $1$): $k\int_{0}^{1}\int_{0}^{y} xy\,dx\,dy=k\int_{0}^{1} y\cdot\tfrac{y^{2}}{2}\,dy=k\int_{0}^{1}\tfrac{y^{3}}{2}\,dy=k\cdot\tfrac{1}{8}$. Set equal to $1$: $\tfrac{k}{8}=1\Rightarrow k=8$.

$f_X(x)=\int_{0}^{1}(x+y)\,dy=\left[xy+\tfrac{y^{2}}{2}\right]_{0}^{1}=x+\tfrac{1}{2}$, for $0<x<1$. Check: $\int_{0}^{1}\left(x+\tfrac{1}{2}\right)dx=\tfrac{1}{2}+\tfrac{1}{2}=1$. Valid.

Hold $y$ fixed; $x$ ranges from $0$ to $y$: $f_Y(y)=\int_{0}^{y} 8xy\,dx=8y\cdot\tfrac{y^{2}}{2}=4y^{3}$, for $0<y<1$. Check: $\int_{0}^{1} 4y^{3}\,dy=1$. Valid.

Differentiate the cdf in both arguments: $f(x,y)=\dfrac{\partial^{2}}{\partial x\,\partial y}F(x,y)$, with $F(x,y)=\Pr(X\leq x,\,Y\leq y)$. If independent, $F(x,y)=F_X(x)F_Y(y)$, so $\Pr(a<X\leq b,\ c<Y\leq d)=\bigl(F_X(b)-F_X(a)\bigr)\bigl(F_Y(d)-F_Y(c)\bigr)$.

Differentiate: $\dfrac{\partial}{\partial x}F=e^{-x}(1-e^{-2y})$, then $\dfrac{\partial}{\partial y}$ gives $f(x,y)=2e^{-x}e^{-2y}$. Since $f$ factors as $\left(e^{-x}\right)\left(2e^{-2y}\right)$ on a rectangular support, $X\sim\text{Exp}(1)$ and $Y\sim\text{Exp}(2)$ are **independent**.

$f_{Y\mid X}(y\mid x)=\dfrac{f(x,y)}{f_X(x)}$, defined where $f_X(x)>0$. For each fixed $x$, this is a genuine density in $y$: it is nonnegative and integrates to $1$ over $y$.

The marginal is $f_Y(y)=4y^{3}$ on $0<y<1$. $f_{X\mid Y}(x\mid y)=\dfrac{8xy}{4y^{3}}=\dfrac{2x}{y^{2}}$, for $0<x<y$. This is the density of $X$ given $Y=y$; it integrates to $1$ since $\int_{0}^{y}\tfrac{2x}{y^{2}}\,dx=1$.

$\Pr(X=1\mid Y=2)=\dfrac{\Pr(X=1,Y=2)}{\Pr(Y=2)}=\dfrac{0.15}{0.30}=0.5$.

Conditional density of $X$ given $Y=y$: $f_{X\mid Y}(x\mid y)=\dfrac{2}{f_Y(y)}$, where $f_Y(y)=\int_{0}^{y}2\,dx=2y$. So $f_{X\mid Y}(x\mid y)=\tfrac{1}{y}$, i.e. $X\mid Y=y$ is uniform on $(0,y)$. At $y=\tfrac{1}{2}$: $\Pr\!\left(X<\tfrac{1}{4}\mid Y=\tfrac{1}{2}\right)=\dfrac{1/4}{1/2}=\tfrac{1}{2}$.

$E[X\mid Y=y]=\int_{0}^{y} x\cdot\dfrac{2x}{y^{2}}\,dx=\dfrac{2}{y^{2}}\int_{0}^{y} x^{2}\,dx=\dfrac{2}{y^{2}}\cdot\dfrac{y^{3}}{3}=\dfrac{2y}{3}$. So $E[X\mid Y]=\tfrac{2}{3}Y$ — given $Y=y$, $X$ averages two-thirds of the way to $y$.

Given the sum $X+Y=t$, the conditional distribution of $X$ is uniform on $(0,t)$ (a known property of two i.i.d. exponentials). Therefore $E[X\mid X+Y=t]=\tfrac{t}{2}$. By symmetry each of $X,Y$ carries half the total on average.

$X$ and $Y$ are independent iff $f(x,y)=f_X(x)\,f_Y(y)$ for all $x,y$. Equivalently the joint pdf factors into a function of $x$ alone times a function of $y$ alone **and** the support is a rectangle (a product set).

The support is the unit square (rectangular). The marginals are $f_X(x)=\int_{0}^{1}6xy^{2}\,dy=2x$ and $f_Y(y)=\int_{0}^{1}6xy^{2}\,dx=3y^{2}$. Product: $f_X(x)f_Y(y)=6xy^{2}=f(x,y)$. **Yes, independent.**

**No.** The support is triangular ($x<y$), not a rectangle, so the range of $X$ depends on $Y$. Support coupling alone breaks independence. Concretely $f_X(x)=4x(1-x^{2})$ and $f_Y(y)=4y^{3}$; their product is not $8xy$. A factorable-looking density on a non-rectangular region is a classic independence trap.

Independence factors the expectation: $E[g(X)\,h(Y)]=E[g(X)]\,E[h(Y)]$. In particular $E[XY]=E[X]E[Y]$, which forces $\operatorname{Cov}(X,Y)=0$. It also lets you compute the MGF of a sum as a product of MGFs.

Factor by independence: $E[X^{2}Y]=E[X^{2}]E[Y]=5\cdot 3=15$. $E[XY]=E[X]E[Y]=6$ and $E[(XY)^{2}]=E[X^{2}]E[Y^{2}]=5\cdot 10=50$. $\operatorname{Var}(XY)=50-6^{2}=50-36=14$.

Joint density $f(x,y)=e^{-x}e^{-y}$ on $x,y>0$. Integrate over $x+y\leq 1$: $\int_{0}^{1}\int_{0}^{1-x} e^{-x}e^{-y}\,dy\,dx=\int_{0}^{1} e^{-x}\left(1-e^{-(1-x)}\right)dx$. $=\int_{0}^{1}\left(e^{-x}-e^{-1}\right)dx=(1-e^{-1})-e^{-1}=1-2e^{-1}\approx 0.264$.

For the uniform on the unit square $f(x,y)=1$, so the probability equals the area where $y>x^{2}$. Area $=\int_{0}^{1}\left(1-x^{2}\right)dx=1-\tfrac{1}{3}=\tfrac{2}{3}$. Thus $\Pr(Y>X^{2})=\tfrac{2}{3}$.

By symmetry of $f$ in $x$ and $y$, $\Pr(X<Y)=\Pr(X>Y)$, and the diagonal has probability $0$, so each is $\tfrac{1}{2}$. Directly: $\int_{0}^{1}\int_{0}^{y}(x+y)\,dx\,dy=\int_{0}^{1}\left(\tfrac{y^{2}}{2}+y^{2}\right)dy=\int_{0}^{1}\tfrac{3y^{2}}{2}\,dy=\tfrac{1}{2}$.

The region is $\{x>0,\,y>0,\,x+y<2\}$, a triangle of area $\tfrac{1}{2}(2)(2)=2$. Uniform means constant density equal to $1/\text{area}$: $f(x,y)=\tfrac{1}{2}$ on the triangle, and $0$ elsewhere.

Triangle area $=\tfrac{1}{2}$, so $f(x,y)=2$. For fixed $x$, $y$ runs from $x$ to $1$: $f_X(x)=\int_{x}^{1} 2\,dy=2(1-x)$. $E[X]=\int_{0}^{1} x\cdot 2(1-x)\,dx=2\left(\tfrac{1}{2}-\tfrac{1}{3}\right)=2\cdot\tfrac{1}{6}=\tfrac{1}{3}$.

By symmetry, $E[|X-Y|]=2\,E[(X-Y)\mathbf{1}\{X>Y\}]=2\int_{0}^{1}\int_{0}^{x}(x-y)\,dy\,dx$. Inner: $\int_{0}^{x}(x-y)\,dy=\tfrac{x^{2}}{2}$. Outer: $2\int_{0}^{1}\tfrac{x^{2}}{2}\,dx=2\cdot\tfrac{1}{6}=\tfrac{1}{3}$. The expected gap between two independent uniforms is $\tfrac{1}{3}$.

$\operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y]=\tfrac{1}{3}-\left(\tfrac{7}{12}\right)^{2}=\tfrac{1}{3}-\tfrac{49}{144}$. $=\tfrac{48}{144}-\tfrac{49}{144}=-\tfrac{1}{144}\approx -0.00694$. The small negative covariance reflects that $X$ and $Y$ are mildly negatively associated here.

Here $E[X]=0$ and $E[XY]=E[X^{3}]=0$ (by symmetry), so $\operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y]=0$. But $Y$ is completely determined by $X$ ($Y=X^{2}$), so they are clearly **dependent**. Zero covariance only rules out *linear* association.

Each count is binomial: $X_i\sim\text{Bin}(n,p_i)$, so $E[X_i]=np_i$ and $\operatorname{Var}(X_i)=np_i(1-p_i)$. Different categories compete for the same $n$ trials, so they are **negatively** correlated: $\operatorname{Cov}(X_i,X_j)=-np_i p_j$ for $i\neq j$. The negative sign is intuitive — every trial in category $i$ is a trial that cannot land in category $j$.

Marginal $X_1\sim\text{Bin}(10,0.5)$: $\operatorname{Var}(X_1)=np_1(1-p_1)=10(0.5)(0.5)=2.5$. Cross-covariance: $\operatorname{Cov}(X_1,X_2)=-np_1 p_2=-10(0.5)(0.3)=-1.5$. Negative, as expected — more low-risk policies means fewer slots left for medium-risk ones.

Parameters: the two means $\mu_X,\mu_Y$, the two standard deviations $\sigma_X,\sigma_Y$, and the correlation $\rho$ (with $-1<\rho<1$). Key property: every linear combination $aX+bY$ is itself **(univariate) normal**, with mean $a\mu_X+b\mu_Y$ and variance $a^{2}\sigma_X^{2}+b^{2}\sigma_Y^{2}+2ab\,\rho\sigma_X\sigma_Y$. Both marginals are normal: $X\sim N(\mu_X,\sigma_X^{2})$ and $Y\sim N(\mu_Y,\sigma_Y^{2})$.

The conditional distribution is normal with the regression-line mean $E[Y\mid X=x]=\mu_Y+\rho\dfrac{\sigma_Y}{\sigma_X}(x-\mu_X)$ and constant variance $\operatorname{Var}(Y\mid X=x)=\sigma_Y^{2}(1-\rho^{2})$. Here $E[Y\mid X=12]=20+0.6\cdot\dfrac{5}{2}(12-10)=20+0.6(2.5)(2)=23$. $\operatorname{Var}(Y\mid X=12)=5^{2}(1-0.6^{2})=25(0.64)=16$, so $\operatorname{SD}=4$ (note: it does not depend on $x$).

Set $\rho=0$ in the joint density: the cross term vanishes and the density factors as $f(x,y)=f_X(x)\,f_Y(y)$ (a product of two univariate normals), so $X\perp Y$. Conversely independence always gives $\rho=0$. Hence **for the bivariate normal, $\rho=0\iff$ independence.** This equivalence is **special to the bivariate normal**. In general $\rho=0$ (zero covariance) does *not* imply independence — e.g. $X$ uniform on $\{-1,0,1\}$ with $Y=X^{2}$ has $\rho=0$ yet $Y$ is a function of $X$.

$\operatorname{Var}(X-Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)-2\operatorname{Cov}(X,Y)$. $=4+9-2(-2)=4+9+4=17$. Note the sign: subtracting a variable still adds its variance, and $-2\operatorname{Cov}$ here becomes $+4$.

Independent, so variances add: $\operatorname{Var}(S)=3\cdot 50^{2}=3\cdot 2500=7500$. Standard deviation: $\operatorname{SD}(S)=\sqrt{7500}\approx 86.6$. (The mean is $3\cdot 200=600$, but SD grows like $\sqrt{n}$, not $n$.)

First recover the covariance: $\operatorname{Cov}(X,Y)=\rho\,\sigma_X\sigma_Y=0.5\cdot 2\cdot 3=3$. $\operatorname{Var}(X+Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)+2\operatorname{Cov}(X,Y)=4+9+2(3)=19$.

$\operatorname{Var}\!\left(\sum X_i\right)=\sum\operatorname{Var}(X_i)+2\sum_{i<j}\operatorname{Cov}(X_i,X_j)$. There are $3$ variance terms and $\binom{3}{2}=3$ covariance pairs. $=3(10)+2(3)(4)=30+24=54$.

$\operatorname{Cov}(X,Y)=\rho\sigma_X\sigma_Y=-0.4(10)(20)=-80$. $\operatorname{Var}(X+Y)=10^{2}+20^{2}+2(-80)=100+400-160=340$. $\operatorname{SD}(X+Y)=\sqrt{340}\approx 18.4$. The negative correlation pulls total variance below the independent-case value of $500$.

$E[X]=E\big[E[X\mid Y]\big]$ — average the conditional mean over the conditioning variable. Worked use: if $E[X\mid Y]=2Y$ and $E[Y]=5$, then $E[X]=E[2Y]=2\,E[Y]=2(5)=10$. Note the inner expectation $E[X\mid Y]$ is a random variable (a function of $Y$); the outer one averages it out.

Given $N=n$, $H\mid N=n$ is binomial$(n,0.4)$, so $E[H\mid N]=0.4N$. Law of total expectation: $E[H]=E[0.4N]=0.4\,E[N]=0.4\cdot 5=2$.

$\operatorname{Var}(Y)=E\!\left[\operatorname{Var}(Y\mid X)\right]+\operatorname{Var}\!\left(E[Y\mid X]\right)$. The two pieces are the *mean of the conditional variance* plus the *variance of the conditional mean*. Forgetting either term is the standard error.

Poisson: $E[N\mid\Lambda]=\Lambda$ and $\operatorname{Var}(N\mid\Lambda)=\Lambda$. $E[N]=E[\Lambda]=3$. EVVE: $\operatorname{Var}(N)=E[\operatorname{Var}(N\mid\Lambda)]+\operatorname{Var}(E[N\mid\Lambda])=E[\Lambda]+\operatorname{Var}(\Lambda)=3+9=12$.

Conditional pieces: $E[Y\mid X]=\tfrac{X}{2}$ and $\operatorname{Var}(Y\mid X)=\tfrac{X^{2}}{12}$. $X\sim U(0,6)$: $E[X]=3$, $\operatorname{Var}(X)=3$, $E[X^{2}]=3+9=12$. $E[\operatorname{Var}(Y\mid X)]=\tfrac{E[X^{2}]}{12}=\tfrac{12}{12}=1$; $\operatorname{Var}(E[Y\mid X])=\tfrac{1}{4}\operatorname{Var}(X)=\tfrac{3}{4}$. $\operatorname{Var}(Y)=1+\tfrac{3}{4}=\tfrac{7}{4}=1.75$.

EVVE: $\operatorname{Var}(S)=E[\operatorname{Var}(S\mid\Theta)]+\operatorname{Var}(E[S\mid\Theta])$. First term: $E[400\Theta]=400\cdot 2=800$. Second term: $\operatorname{Var}(100\Theta)=100^{2}\cdot 0.5=10000\cdot 0.5=5000$. $\operatorname{Var}(S)=800+5000=5800$.

$E[H\mid N]=0.4N$ and $\operatorname{Var}(H\mid N)=N(0.4)(0.6)=0.24N$. EVVE: $\operatorname{Var}(H)=E[0.24N]+\operatorname{Var}(0.4N)=0.24\,E[N]+0.16\,\operatorname{Var}(N)$. $=0.24(5)+0.16(5)=1.2+0.8=2$. (Indeed thinning a Poisson gives $H\sim\text{Poisson}(2)$, variance $2$ — consistent.)