Question

Question:Suppose thatXandYare random variables such that(X, Y)must belong to the rectangle in thexy-plane containing all points(x, y)for which 0≤x≤3 and 0≤y≤4. Suppose also that the joint c.d.f. ofXandYat every point

(x,y) in this rectangle is specified as follows:

\({\bf{F}}\left( {{\bf{x,y}}} \right){\bf{ = }}\frac{{\bf{1}}}{{{\bf{156}}}}{\bf{xy}}\left( {{{\bf{x}}^{\bf{2}}}{\bf{ + y}}} \right)\)

Determine

(a) Pr(1≤X≤2 and 1≤Y≤2);

(b) Pr(2≤X≤4 and 2≤Y≤4);

(c) the c.d.f. ofY;

(d) the joint p.d.f. ofXandY;

(e) Pr(Y≤X).

Short answer

1. \(\Pr \left( {1 \le X \le 2\;and\;1 \le y \le 2} \right) = \frac{{10}}{{156}}\)
2. \(\Pr \left( {2 \le X \le 4\;and\;2 \le y \le 4} \right) = \frac{{50}}{{156}}\)
3. The c.d.f of Y is\({F_Y}\left( y \right) = \left\{ \begin{array}{l}0\;if\;y < 0\\\frac{1}{{52}}y\left( {9 + y} \right)\;if\;0 \le y \le 4\\0\;if\;y > 1\end{array} \right.\)
4. The X and Y joint is\(f\left( {x,y} \right) = \frac{1}{{156}}\left( {3{x^2} + 2x} \right){I_{\left( {0,3} \right) \times \left( {0,4} \right)}}\left( {x,y} \right)\)
5. \(\Pr \left( {Y \le X} \right) = 0.447\)

Step-by-step solution

Given information

Xand Yare two random variables. The point (X, Y) belongs to the rectangle in the xy-plane. All the points of (x,y) contain in the plane. The range of xand yis,\(0 \le x \le 3\)and\(0 \le y \le 4\)respectively.

The joint cumulative density function of Xand Yat every point in the rectangle is,

\(F\left( {x,y} \right) = \frac{1}{{156}}xy\left( {{x^2} + y} \right)\)

Determine the probability

a.

We have to calculate the probability,

\(\begin{array}{c}\Pr \left( {1 \le X \le 2,1 \le Y \le 2} \right) = \Pr \left( {X \le 2,1 \le y \le 2} \right) - \Pr \left( {X \le 1,1 \le y \le 2} \right)\\ = \left( \begin{array}{l}\left( {\Pr \left( {X \le 2,Y \le 2} \right) - \Pr \left( {X \le 2,Y \le 1} \right)} \right)\\ - \left( {\Pr \left( {X \le 1,Y \le 2} \right) - \Pr \left( {X \le 1,Y \le 1} \right)} \right)\end{array} \right)\\ = \left( {F\left( {2,2} \right) - F\left( {2,1} \right)} \right) - \left( {F\left( {1,2} \right) - F\left( {1,1} \right)} \right)\\ = \frac{1}{{156}}\left( {\left| {24 - 10} \right| - \left| {6 - 2} \right|} \right)\end{array}\)

\( = \frac{{10}}{{156}}\)

Thus, the probability is \(\frac{{10}}{{156}}\) \(\)

Determine the probability

b.

We have to calculate the probability,

\(\begin{array}{c}\Pr \left( {2 \le X \le 4,2 \le Y \le 4} \right) = \Pr \left( {2 \le X \le 3,2 \le Y \le 4} \right)\\ = \left( {F\left( {3,4} \right) - F\left( {3,2} \right)} \right) - \left( {F\left( {1,2} \right) - F\left( {1,1} \right)} \right)\\ = \frac{1}{{156}}\left( {\left| {156 - 66} \right| - \left| {64 - 24} \right|} \right)\\ = \frac{{50}}{{156}}\end{array}\)

Thus, the probability is \(\frac{{50}}{{156}}\)

Determine the cumulative Density function

c.

The c.d.f of Yis

\(\begin{array}{c}{F_Y}\left( y \right) = \mathop {\lim }\limits_{x \to \infty } {F_{X,Y}}\left( {3,y} \right)\\ = {\left. {\frac{1}{{156}}xy\left( {{x^2} + y} \right)} \right|_{x = 3}}\\ = \frac{1}{{156}}3y\left( {{3^2} + y} \right)\\ = \frac{1}{{52}}y\left( {9 + y} \right)\end{array}\)

Now, in general, we can say that, \({{\bf{F}}_{\bf{Y}}}\left( {\bf{y}} \right){\bf{ = }}\left\{ \begin{array}{l}{\bf{0}}\;{\bf{if}}\;{\bf{y < 0}}\\\frac{{\bf{1}}}{{{\bf{52}}}}{\bf{y}}\left( {{\bf{9 + y}}} \right)\;{\bf{if}}\;{\bf{0}} \le {\bf{y}} \le {\bf{4}}\\{\bf{0}}\;{\bf{if}}\;{\bf{y > 1}}\end{array} \right.\)

Determine the joint Probability Density Function

d.

As there is a rectangle in the xy-plane, the boundary of the rectangular set can be excluded for the sharp corner points. So,\(F\left( {x,Y} \right)\)it is not differentiable at those points.

Now for\(0 < x < 3\)and\(0 < y < 4\),

\(\begin{array}{c}{f_{X,Y}}\left( {x,y} \right) = \frac{{{\partial ^2}F\left( {x,y} \right)}}{{\partial x\partial y}}\\ = \frac{\partial }{{\partial x}}\left( {\frac{\partial }{{\partial y}}\frac{1}{{156}}xy\left( {{x^2} + y} \right)} \right)\\ = \frac{1}{{156}}\frac{\partial }{{\partial x}}\left( {{x^3} + 2xy} \right)\\ = \frac{1}{{156}}\left( {3{x^2} + 2y} \right)\end{array}\)

Now for\(\left( {x,y} \right) \in {\mathbb{R}^2}:\left( {0,3} \right) \times \left( {0,4} \right)\)we can say that,\({f_{X,Y}}\left( {x,y} \right) = 0\).

For \(F\left( {x,y} \right) = 0\;or\;1\) and we set \(f\left( {x,y} \right) = 0\) at the boundary, the joint p.d.f of X and Y is given by,

\({\bf{f}}\left( {{\bf{x,y}}} \right){\bf{ = }}\frac{{\bf{1}}}{{{\bf{156}}}}\left( {{\bf{3}}{{\bf{x}}^{\bf{2}}}{\bf{ + 2x}}} \right){{\bf{I}}_{\left( {{\bf{0,3}}} \right){\bf{ \times }}\left( {{\bf{0,4}}} \right)}}\left( {{\bf{x,y}}} \right)\)

Determine the probability

e.

We have to calculate the probability,

\(\begin{array}{c}\Pr \left( {Y \le X} \right) = \int\limits_0^3 {\int\limits_0^x {\frac{1}{{156}}\left( {3{x^2} + 2x} \right)dxdy} } \\ = \frac{1}{{156}}\int\limits_0^3 {\left. {\left( {3{x^2}y + {y^2}} \right)} \right|_0^xdx} \\ = \frac{1}{{156}}\int\limits_0^3 {\left( {3{x^3} + {x^2}} \right)dx} \end{array}\)

\(\begin{array}{c} = \frac{1}{{156}}\left. {\left( {\frac{{3{x^4}}}{4} + \frac{{{x^3}}}{3}} \right)} \right|_0^3\\ = \frac{1}{{156}}\left( {\frac{{243}}{4} + 9} \right)\\ = 0.447\end{array}\)

Thus, the probability is 0.447