Question

The joint probability density function of random variables X and Y is given by f(x,y) ={10xy^2 0≤x≤y≤1,0 otherwise.

(a) Compute the conditional probability f_X|Y(x|y).

(b) Compute E(Y) and P(Y >1/2).

(c) Let W=X/Y. Compute the density function of W.

(d) Are X and Y independent? Justify briefly.

Answer 1

(a)    $\int_{s}\int_{t}f_{X,Y}(s,t)dt \ ds = 1$

$\int_{s}\int_{t}f_{X,Y}(s,t)dt \ ds =\int_{s}\left ( \int_{t}cs \ dt \right )\ ds = 1$

$\Rightarrow \int_{0}^{1}\int_{0}^{2}cs \ dt \ ds = \int_{0}^{1} \left [ cst \right ]_{0}^{2} ds = \int_{0}^{1}\left ( 2cs \right )ds = 2\left [\frac{cs^{2}}{2} \right ]_{0}^{1} = 1$

$\Rightarrow \left [cs^{2} \right ]_{0}^{1} = c(1-0) = 1$

$\Rightarrow c = 1$

$\Rightarrow f_{X,Y}(s,t)=s$

(b) the marginal probability density function of Y: Also, substituing the value of c,

$f_{Y}(t) = \int_{s}f_{X,Y}(s,t)ds = \int_{0}^{1}s \ ds =\left [\frac{s^{2}}{2} \right ] _{0}^{1} = \frac{1}{2} - 0 = \frac{1}{2}$

(c) the conditional probabiltiy density function of X given Y=1:

$f_{X|Y=1}(s,t) =\frac{f_{X,Y}(s,t)}{f_{Y}(t)} = \frac{s}{\frac{1}{2}} = 2s \ \ 0< s< 1$

(d)Pr {X=Y}: Setting limits of Y from 0 to X i.e., 0 < t < s, we get

$f_{X=Y}(s,t) =\int_{0}^{1}\left [ \int_{0}^{s} s \ dt \right ] ds = \int_{0}^{1} s \ t_{0}^{s} \ ds = \int_{0}^{1} s^{2} \ ds = \left [ \frac{s^{2}}{2} \right ]_{0}^{1} = \frac{1}{2}$

(e) Pr{X+Y<1} = Pr (Y < 1-X )

$f_{X+Y<1}(s,t) =\int_{0}^{1}\left [ \int_{0}^{1-s} s \ dt \right ] ds = \int_{0}^{1} s \ t_{0}^{1 - s} \ ds = \int_{0}^{1} s(1 - s) \ ds = \int_{0}^{1} s-s^{2} \ ds = \left [ \frac{s^{2}}{2} - \frac{s^{3}}{3} \right ]_{0}^{1} = \frac{1}{2} - \frac{1}{3} = \frac{3-2}{6} =\frac{1}{6}$

(f) check whether X and Y are independent:

$f_{Y}(t) = \frac{f_{Y|X=s}(t,s)}{f_{X}(s)} =\frac{f_{X,Y}(s,t)}{f_{X}(s)}$

If conditional is equivalent to marginal then X and Y are independent.

i.e. ${\color{DarkBlue} f_{X,Y} (s,t)=f_{X}(s) f_{Y}(t)}$

We already found f(Y), Now we need to find f(X):

$f_{X}(s) = \int_{t}f_{X,Y}(s,t)dt = \int_{0}^{2}s \ dt = s\left [ t \right ]_{0}^{2} = 2s \ where \ 0<s<1$

$\Rightarrow f_{Y}(t) = \frac{1}{2} \ and \ f_{X}(s) = 2s$

$\Rightarrow f_{Y}(t)\times \ f_{X}(s) = \frac{1}{2} \times 2s = s$

$\Rightarrow f_{Y}(t) \ f_{X}(s) = f_{X,Y}(s,t)$

Hence X and Y are independent.