Bivariate Distributions of the Continuous Type

(a) We have

$$\begin{aligned}ℙ(\{(𝑋,𝑌)\in 𝐴\})={\iint }_{𝐴}𝑓(𝑥,𝑦)\, \mathrm{d}𝑥\, \mathrm{d}𝑦 & ={\int }_{0}^{1}{\int }_{0}^{𝑥}\frac{3}{2}{𝑥}^{2}(1-|𝑦|)\, \mathrm{d}𝑦\, \mathrm{d}𝑥 \\ & ={\int }_{0}^{1}\frac{3}{2}{𝑥}^{2}{\int }_{0}^{𝑥}1-𝑦\, \mathrm{d}𝑦\, \mathrm{d}𝑥 \\ & ={\int }_{0}^{1}\frac{3}{2}{𝑥}^{2}{\left[𝑦-\frac{{𝑦}^{2}}{2}\right]}_{0}^{𝑥}\, \mathrm{d}𝑥 \\ & =\frac{3}{2}{\int }_{0}^{1}{𝑥}^{3}-\frac{{𝑥}^{4}}{2}\, \mathrm{d}𝑥 \\ & =\frac{9}{40}.\end{aligned}$$

(b) The marginal PDF of $𝑋$ is obtained by integrating out $𝑦$:

$$\begin{aligned}{𝑓}_{𝑋}(𝑥) & ={\int }_{-1}^{1}\frac{3}{2}{𝑥}^{2}(1-|𝑦|)\, \mathrm{d}𝑦 \\ & =\frac{3}{2}{𝑥}^{2}{\int }_{-1}^{1}1-|𝑦|\, \mathrm{d}𝑦 \\ & =\frac{3}{2}{𝑥}^{2}\cdot 2{\int }_{0}^{1}1-|𝑦|\, \mathrm{d}𝑦 \\ & =\frac{3}{2}{𝑥}^{2}\cdot 2{\left[𝑦-\frac{{𝑦}^{2}}{2}\right]}_{0}^{1} \\ & =\frac{3}{2}{𝑥}^{2}.\end{aligned}$$

Now, we integrate out $𝑥$ to get the marginal PDF of $𝑌$:

$$\begin{aligned}{𝑓}_{𝑌}(𝑦) & ={\int }_{-1}^{1}\frac{3}{2}{𝑥}^{2}(1-|𝑦|)\, \mathrm{d}𝑥 \\ & =\frac{3}{2}(1-|𝑦|){\int }_{-1}^{1}{𝑥}^{2}\, \mathrm{d}𝑥 \\ & =\frac{3}{2}(1-|𝑦|)\cdot {\left[\frac{{𝑥}^{3}}{3}\right]}_{-1}^{1} \\ & =1-|𝑦|.\end{aligned}$$

(c) We have

$${\mu }_{𝑋}={\int }_{-1}^{1}𝑥{𝑓}_{𝑋}(𝑥)\, \mathrm{d}𝑥.$$

Note that ${𝑓}_{𝑋}(𝑥)$ is an even function, so $𝑥{𝑓}_{𝑋}(𝑥)$ is odd; hence integrating it over a symmetric interval gives 0. Thus, ${\mu }_{𝑋}=0$. Similarly, since

$${\mu }_{𝑌}={\int }_{-1}^{1}𝑦{𝑓}_{𝑌}(𝑦)\, \mathrm{d}𝑦.$$

and ${𝑓}_{𝑌}(𝑦)$ is also an even function, we have ${\mu }_{𝑌}=0$.

We now compute ${\sigma }_{𝑋}^{2}$ and ${\sigma }_{𝑌}^{2}$. We have

$$\begin{aligned}{\sigma }_{𝑋}^{2}=\mathbb{E}[{𝑋}^{2}]-{\mu }_{𝑋}^{2}=\mathbb{E}[{𝑋}^{2}] & ={\int }_{-1}^{1}{𝑥}^{2}{𝑓}_{𝑋}(𝑥)\, \mathrm{d}𝑥 \\ & ={\int }_{-1}^{1}{𝑥}^{2}\cdot \frac{3}{2}{𝑥}^{2}\, \mathrm{d}𝑥 \\ & ={\left[\frac{3{𝑥}^{5}}{10}\right]}_{-1}^{1} \\ & =\frac{3}{5}.\end{aligned}$$

To compute ${\sigma }_{𝑌}^{2}$, we have

$$\begin{aligned}{\sigma }_{𝑌}^{2}=\mathbb{E}[{𝑌}^{2}]-{\mu }_{𝑌}^{2}=\mathbb{E}[{𝑌}^{2}] & ={\int }_{-1}^{1}{𝑦}^{2}{𝑓}_{𝑌}(𝑦)\, \mathrm{d}𝑦 \\ & ={\int }_{-1}^{1}{𝑦}^{2}\cdot (1-|𝑦|)\, \mathrm{d}𝑦 \\ & =2{\int }_{0}^{1}{𝑦}^{2}\cdot (1-𝑦)\, \mathrm{d}𝑦 \\ & ={\left[2\left(\frac{{𝑦}^{3}}{3}-\frac{{𝑦}^{4}}{4}\right)\right]}_{0}^{1} \\ & =\frac{1}{6}.\end{aligned}$$

(d) We could compute

$$\operatorname{Cov}(𝑋,𝑌)=\mathbb{E}[𝑋𝑌]-{\mu }_{𝑋}{\mu }_{𝑌}=\mathbb{E}[𝑋𝑌]={\int }_{-1}^{1}{\int }_{-1}^{1}𝑥𝑦𝑓(𝑥,𝑦)\, \mathrm{d}𝑥\, \mathrm{d}𝑦.$$

But we note that, by one glance, $𝑋$ and $𝑌$ are independent since $𝑓(𝑥,𝑦)={𝑓}_{𝑋}(𝑥){𝑓}_{𝑌}(𝑦)$. (In fact, we have not defined the joint PDF of continuous type random variables yet. We will do so right after this exercise.) Therefore, $\operatorname{Cov}(𝑋,𝑌)=0$ and hence $\rho =0$.

(a) We first find ${𝑓}_{𝑋}(𝑥)$. Let $𝑋=𝑥$ be fixed. Imagine a vertical line $𝑥=𝑥$ passing through the support $𝑆$ in Figure 4.3 along the direction of the $𝑦$-axis. Note that it enters $𝑆$ at $(𝑥,𝑥)$ and leaves at $(𝑥,1)$. Integrating $𝑓(𝑥,𝑦)$ over $𝑦$ from $𝑥$ to $1$ gives the marginal PDF of $𝑋$:

$${𝑓}_{𝑋}(𝑥)={\int }_{𝑥}^{1}2\, \mathrm{d}𝑦=2(1-𝑥).$$

Similarly, we can find ${𝑓}_{𝑌}(𝑦)$ by integrating $𝑓(𝑥,𝑦)$ over $𝑥$ from $0$ to $𝑦$.

$${𝑓}_{𝑌}(𝑦)={\int }_{0}^{𝑦}2\, \mathrm{d}𝑥=2𝑦.$$

(b) The conditional PDF of $𝑌$ given $𝑋=𝑥$ ($0\leq 𝑥<1$) is given by

$$h(𝑦|𝑥)=\frac{𝑓(𝑥,𝑦)}{{𝑓}_{𝑋}(𝑥)}=\frac{1}{1-𝑥},\quad 𝑦\in [𝑥,1].$$

That is, given $𝑋=𝑥$, $𝑌$ is uniformly distributed over the interval $[𝑥,1]$, which agrees with our intuition. Recall that if a random variable $𝑊∼𝒰(𝑎,𝑏)$, then its mean and variance are ${\mu }_{𝑊}=(𝑎+𝑏)/2$ and ${\sigma }_{𝑊}^{2}={(𝑏-𝑎)}^{2}/12$, respectively. Therefore, before we compute ${\mu }_{𝑌|𝑥}$ and ${\sigma }_{𝑌|𝑥}^{2}$ using the definition of conditional expectations, we know that ${\mu }_{𝑌|𝑥}=(𝑥+1)/2$ and ${\sigma }_{𝑌|𝑥}^{2}={(1-𝑥)}^{2}/12$. In the following, we will verify this by computing integrals.

(c) ${\mu }_{𝑌|𝑥}$ is given by

$${\mu }_{𝑌|𝑥}={\int }_{𝑥}^{1}𝑦h(𝑦|𝑥)\, \mathrm{d}𝑦={\int }_{𝑥}^{1}\frac{𝑦}{1-𝑥}\, \mathrm{d}𝑦=\frac{1+𝑥}{2}.$$

(d) We need to compute $\mathbb{E}[{𝑌}^{2}|𝑋=𝑥]$. We have

$$\mathbb{E}[{𝑌}^{2}|𝑋=𝑥]={\int }_{𝑥}^{1}{𝑦}^{2}h(𝑦|𝑥)\, \mathrm{d}𝑦={\int }_{𝑥}^{1}\frac{{𝑦}^{2}}{1-𝑥}\, \mathrm{d}𝑦=\frac{1+𝑥+{𝑥}^{2}}{3}.$$

Therefore,

$${\sigma }_{𝑌|𝑥}^{2}=\mathbb{E}[{𝑌}^{2}|𝑋=𝑥]-{\mu }_{𝑌|𝑥}^{2}=\frac{1+𝑥+{𝑥}^{2}}{3}-{\left(\frac{1+𝑥}{2}\right)}^{2}=\frac{{(1-𝑥)}^{2}}{12}.$$