Exam P — Multivariate Distributions Practice Flashcards

**Answer: (C).** Require the density to integrate to $1$ over the rectangle: $\int_{0}^{2}\int_{0}^{1} c\,x^{2}y\,dy\,dx=1$. Inner integral: $\int_{0}^{1} y\,dy=\dfrac{1}{2}$, so we have $c\int_{0}^{2}\dfrac{x^{2}}{2}\,dx$. Outer: $\int_{0}^{2} x^{2}\,dx=\dfrac{8}{3}$, giving $c\cdot\dfrac{1}{2}\cdot\dfrac{8}{3}=\dfrac{4c}{3}$. Set $\dfrac{4c}{3}=1\Rightarrow c=\dfrac{3}{4}$. (Distractor $\tfrac{3}{8}$ comes from forgetting the factor $2$ from the $x$-range; $\tfrac{3}{2}$ from dropping the $\tfrac12$ in the $y$-integral.)

**Answer: (D).** First confirm the density is valid: $\int_{0}^{1}\int_{0}^{2}\dfrac{1}{5}(x+2y)\,dy\,dx=\dfrac{1}{5}\int_{0}^{1}(2x+4)\,dx=\dfrac{1}{5}(1+4)=1.$ Good. Integrate out $y$ over $0<y<2$: $f_X(x)=\int_{0}^{2}\dfrac{1}{5}(x+2y)\,dy=\dfrac{1}{5}\left[xy+y^{2}\right]_{0}^{2}=\dfrac{1}{5}(2x+4)=\dfrac{2}{5}(x+2),\quad 0<x<1.$ Check: $\int_{0}^{1}\dfrac{2}{5}(x+2)\,dx=\dfrac{2}{5}\left(\dfrac{1}{2}+2\right)=\dfrac{2}{5}\cdot\dfrac{5}{2}=1.$ Valid. (Distractor (A) evaluates $y^{2}$ at $1$ instead of $2$; (B) keeps the $\tfrac15$ but mis-evaluates the bracket.)

**Answer: (D).** For fixed $y$, $x$ ranges over $0<x<y$, so the marginal of $Y$ is $f_Y(y)=\int_{0}^{y} 6(1-y)\,dx=6y(1-y)$, for $0<y<1$. Then $\Pr\!\left(Y<\tfrac12\right)=\int_{0}^{1/2} 6y(1-y)\,dy=6\left[\dfrac{y^{2}}{2}-\dfrac{y^{3}}{3}\right]_{0}^{1/2}=6\left(\dfrac{1}{8}-\dfrac{1}{24}\right).$ $=6\cdot\dfrac{3-1}{24}=6\cdot\dfrac{2}{24}=6\cdot\dfrac{1}{12}=\dfrac{1}{2}.$ (Distractor $\tfrac38$ forgets to subtract the $y^3/3$ term; $\tfrac18$ omits the marginal's $6$.)

**Answer: (A).** For a fixed $y>0$, the constraint $0<y<x$ means $x$ ranges from $y$ to $\infty$: $f_Y(y)=\int_{y}^{\infty} e^{-x}\,dx=\left[-e^{-x}\right]_{y}^{\infty}=e^{-y}$, for $y>0$. So $Y\sim\text{Exp}(1)$. (Distractor (B) $y e^{-y}$ would arise from integrating $x$ from $0$ to $y$ — the wrong limits; (D) is the cdf, not the density.)

**Answer: (C).** Integrate the density over the triangle $\{x>0,y>0,x+y<2\}$. For fixed $x\in(0,2)$, $y$ runs from $0$ to $2-x$: $\Pr(X+Y<2)=\int_{0}^{2}\int_{0}^{2-x}\dfrac{1}{8}(x+y)\,dy\,dx.$ Inner: $\int_{0}^{2-x}(x+y)\,dy=x(2-x)+\dfrac{(2-x)^{2}}{2}.$ Let $u=2-x$: this equals $x\,u+\dfrac{u^{2}}{2}$ with $x=2-u$. Expand directly: $x(2-x)+\dfrac{(2-x)^2}{2}=2x-x^2+\dfrac{4-4x+x^2}{2}=2x-x^2+2-2x+\dfrac{x^2}{2}=2-\dfrac{x^2}{2}.$ So $\Pr=\dfrac{1}{8}\int_{0}^{2}\left(2-\dfrac{x^{2}}{2}\right)dx=\dfrac{1}{8}\left[2x-\dfrac{x^{3}}{6}\right]_{0}^{2}=\dfrac{1}{8}\left(4-\dfrac{8}{6}\right)=\dfrac{1}{8}\cdot\dfrac{8}{3}=\dfrac{1}{3}.$

**Answer: (B).** The marginal of $X$: for fixed $x$, $y$ runs from $x$ to $1$, so $f_X(x)=\int_{x}^{1} 2\,dy=2(1-x)$, for $0<x<1$. Conditional density: $f_{Y\mid X}(y\mid x)=\dfrac{f(x,y)}{f_X(x)}=\dfrac{2}{2(1-x)}=\dfrac{1}{1-x}$, for $x<y<1$. This says $Y\mid X=x$ is uniform on $(x,1)$, which integrates to $1$. (Distractor (D) forgets to cancel the $2$; (C) uses the wrong marginal.)

**Answer: (D).** The density factors as $f(x,y)=\left(\dfrac{1}{2}\right)\left(3y^{2}\right)$ on the rectangle $0<x<2$, $0<y<1$, so $X\sim U(0,2)$ and $Y$ has density $f_Y(y)=3y^{2}$, and $X,Y$ are independent. Because they are independent, conditioning on $X=1.5$ does not change the distribution of $Y$: $E[Y\mid X=1.5]=E[Y]=\int_{0}^{1} y\cdot 3y^{2}\,dy=\int_{0}^{1}3y^{3}\,dy=\left[\dfrac{3y^{4}}{4}\right]_{0}^{1}=\dfrac{3}{4}.$ (Distractor (C) $\tfrac23$ uses $f_Y=2y$ by mistake; (A) ignores the density shape and guesses $\tfrac12$.)

**Answer: (C).** For fixed $x$, $y$ ranges over $0<y<1-x$. The marginal of $X$: $f_X(x)=\int_{0}^{1-x}24xy\,dy=24x\cdot\dfrac{(1-x)^{2}}{2}=12x(1-x)^{2}.$ Conditional density: $f_{Y\mid X}(y\mid x)=\dfrac{24xy}{12x(1-x)^{2}}=\dfrac{2y}{(1-x)^{2}},\quad 0<y<1-x.$ Then $E[Y\mid X=x]=\int_{0}^{1-x} y\cdot\dfrac{2y}{(1-x)^{2}}\,dy=\dfrac{2}{(1-x)^{2}}\cdot\dfrac{(1-x)^{3}}{3}=\dfrac{2(1-x)}{3}.$ (This is the mean of a Beta-type density on $(0,1-x)$. Distractor (B) treats $Y\mid X$ as uniform on $(0,1-x)$.)