Question

The joint probability density function 𝑓(𝑥,𝑦) for the random variables 𝑋 and 𝑌 is defined as follows

𝑓(𝑥,𝑦)= 3/2

Are the random variables 𝑋 and 𝑌 independent?

Answer 1

Solution:

Assume the given joint pdf is as follows:

Given random variables X and Y have the following joint probability density function:

$f_{X Y}(x, y)=\left\{\begin{array}{rr}\frac{2}{5}(3 x+2 y) ; & 0 \leq x \leq 1,0 \leq y \leq 1 \\ 0 ; & o . w\end{array}\right.$

It is assumed that $f_{X Y}(x, y)$ satisfies the properties of a joint probability density function.

(a)

Required to find the marginal probability density function of X.

The marginal probability density function of X is given as follows:

$\begin{aligned} f_{X}(x) &=\int_{y} f_{X Y}(x, y) d y \\ &=\int_{0}^{1} \frac{2}{5}(3 x+2 y) d y \\ &=\frac{2}{5}\left(3 x y+\frac{2 y^{2}}{2}\right)_{0}^{1} \\ &=\frac{2}{5}(3 x+1-0) \\ &=\frac{2(3 x+1)}{5} \end{aligned}$

The marginal probability density function of Y is given as follows:

$\begin{aligned} f_{Y}(y) &=\int_{x} f_{X Y}(x, y) d x \\ &=\int_{0}^{1} \frac{2}{5}(3 x+2 y) d x \\ &=\frac{2}{5}\left(2 x y+\frac{3 x^{2}}{2}\right)_{0}^{1} \\ &=\frac{2}{5}(2y+\dfrac32) \end{aligned}$

Now, we check whether $P\left(X \leq \dfrac{1}{2}, Y \leq \dfrac{1}{3}\right)=P\left(X \leq \dfrac{1}{2}\right)\left( Y \leq \dfrac{1}{3}\right)$

We take any values to verify this relation.

$\begin{aligned} P\left(X \leq \frac{1}{2}, Y \leq \frac{1}{3}\right) &=\int_{y} \int_{x} f_{X Y}(x, y) d x d y \\ &=\int_{0}^{\frac{1}{3}} \int_{0}^{\frac{1}{2}} \frac{2}{5}(3 x+2 y) d x d y \\ &=\frac{2}{5} \int_{0}^{\frac{1}{3}}\left(\frac{3 x^{2}}{2}+2 x y\right)_{0}^{\frac{1}{2}} d y \\ &=\frac{2}{5} \int_{0}^{\frac{1}{3}}\left(\frac{3\left(\frac{1}{2}\right)^{2}}{2}+2\left(\frac{1}{2}\right) y-0\right) d y \\ &=\frac{2}{5}\left(\frac{3 y}{8}+\frac{y^{2}}{2}\right)_{0}^{\frac{1}{3}} \\ &=\frac{2}{5} \int_{0}^{\frac{1}{3}}\left(\frac{3}{8}+y\right) d y \end{aligned}$

$\begin{aligned} &=\frac{2}{5}\left(\frac{3\left(\frac{1}{3}\right)}{8}+\frac{\left(\frac{1}{3}\right)^{2}}{2}-0\right) \\ &=\frac{2}{5}\left(\frac{1}{8}+\frac{1}{18}\right) \\ &=0.0722 \end{aligned}$

Next, $P\left(X \leq \dfrac{1}{2}\right)\left( Y \leq \dfrac{1}{3}\right)$

$=\dfrac{2(3.\dfrac12+1)}{5}\times\dfrac{2}{5}(2.\dfrac13+\dfrac32) \\=0.866...$

Since, they are unequal, the random variables are not independent.