Question

Answer 1

The two continuous random variables, and
2
, are independent and have the same pdf,

clearly , we can say that and
2
, are IID random variables.

Define the tow variables U and V as follows.

$\\U= X_1 \sqrt{X_2} \ , \ 0 \leq U\leq 1 \\ V= X_2 \sqrt{X_1} \ , \ 0 \leq V\leq 1$

$\\ \Rightarrow X_1= \frac{u^{4/3}}{v^{2/3}} \\ \\ X_2= \frac{v^{4/3}}{u^{2/3}}$

Now , the JACOBIAN of U and V is

$\\ J(U,V) = \begin{vmatrix} \frac{\partial {X_1}}{\partial {u}} & \ \frac{\partial {X_2}}{\partial {u}} \\\\ \frac{\partial {X_1}}{\partial {v}} & \ \frac{\partial {X_2}}{\partial {v}} \end{vmatrix} \\\\\\ = \begin{vmatrix} \frac{4}{3} \frac{u^{1/3}}{v^{2/3}} & \ \frac{-2}{3} \frac{u^{4/3}}{v^{5/3}} \\\\ \frac{-2}{3} \frac{v^{4/3}}{u^{5/3}} & \ \frac{4}{3} \frac{v^{1/3}}{u^{2/3}} \end{vmatrix} \\\\\\ = \frac{4}{3} \frac{1}{(uv)^{1/3}}$

1.

Now , the joint pdf of U and V ,

$f_{U,V}(u,v) \\= f_{X_1}(x_1)*f_{X_2}(x_2)*\left |J(U,V) \right | \\\\= 4[\frac{u^{4/3}}{v^{2/3}}]^{3} * 4[\frac{v^{4/3}}{u^{2/3}}]^{3} * \frac{4}{3} \frac{1}{(uv)^{1/3}} \\\\= (\frac{8}{\sqrt3})^2(uv)^{-5/3} \\\\= (\frac{8}{u^{5/3}\sqrt3})*(\frac{8}{v^{5/3}\sqrt3}) \\\\ = f_U(u)* f_V(v) \\\\ \therefore f_{U,V}(u,v) = (\frac{8}{\sqrt3})^2(uv)^{-5/3}$

2.

Therefore , the marginal PDF of U is ,

$\\f_U(u)= \frac{8}{u^{5/3}\sqrt3}$ where,  $0 \leq u \leq 1$

[consider u=x]

f(x)=[8x^(-5/3)1/3 (0.5) +180 160 140 +120 -100 -80 +60 +40 20 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

3, Therefore , the marginal PDF of V is ,

$\\f_V(v)= \frac{8}{v^{5/3}\sqrt3}$ where, $0 \leq v \leq 1$

[consider v=x]

f(x)=[8x^(-5/3)1/3 (0.5) +180 160 140 +120 -100 -80 +60 +40 20 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

4.

Yes, U and V are independent as the joint PDF of U and V is equal to the multiplication of the marginal PDF of U and V.

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