Question

The joint pdf of a bivariate random variable (X, Y) is
f(x, y) =
4xy, 0 < x < 1,0 < y < 1
,
0, Otherwise
Find P(X + Y < 1).​

Answer 1

Answer:

The $P(X + Y < 1)$ is given by $\frac{4}{9}$

Step-by-step explanation:

Given:

$f(X, Y)= \begin{cases}4 X Y & ; 0 \leq X \leq 1,0 \leq Y \leq 1 \\ 0 & ; \text { otherwise }\end{cases}$

To find: $P(X + Y < 1).$

Step 1:

$\begin{aligned}&E(X Y)=\int_{0}^{1} \int_{0}^{1} X Y .4 X Y \mathrm{~d} x d y \\&E(X Y)=4 \int_{0}^{1} \int_{0}^{1} X^{2} Y^{2} \mathrm{~d} x d y \\&E(X Y)=4 \int_{0}^{1} Y^{2}\left[\frac{X^{3}}{3}\right]_{0}^{1} d y \\&E(X Y)=4 \int_{0}^{1} Y^{2}\left(\frac{1}{3}\right) d y \\&E(X Y)=\frac{4}{3}\left[\frac{Y^{3}}{3}\right]_{0}^{1} \\&E(X Y)=\frac{4}{3}\left(\frac{1}{3}\right) \\&\mathbf{E}(\mathbf{X Y})=\frac{4}{9}\end{aligned}$

Answer 2

Answer:

p(x+y< 1)= $\frac{1}{6}$

Step-by-step explanation:

Given the joint pdf is

$f(x,y)=\left \{ {{4xy ,0 < x < 1, 0 < y < 1} \atop {0, otherwise}} \right.\\ \\ P(x+y < 1)=\int\limits^1_0\int\limits^{1-x}_0 {4xy} \, .dx .dy\\ =\int\limits^1_0 {4x[\frac{y^{2}}{2} ] ^{1-x} _{0} } \, dx$

$=2\int\limits^1_0 {x(1-x)^{2} } \, .dx \\=2\int\limits^1_0 {x-2x^{2} +x^{3} } \, .dx \\\\=2[\frac{x^{2} }{2}-\frac{2x^{3} }{3} +\frac{x^{4} }{4} ]^{1}__{0}$

$2[\frac{1}{2} -\frac{2}{3} +\frac{1}{4} ] \\\\=2[\frac{6-8+3}{12} ] \\=2\frac{1}{12} \\=\frac{1}{6} .$

Therefore,P(x+y<1)= $\frac{1}{6} .$

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