First find the marginal distributions:
[tex]f_X(x)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dy=\int_0^\infty6e^{-x(y+6)}\,\mathrm dy[/tex]
[tex]\implies f_X(x)=\begin{cases}6e^{-6x}&\text{for }x>0\\0&\text{otherwise}\end{cases}[/tex]
[tex]f_Y(y)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dx=\int_0^\infty6e^{-x(y+6)}\,\mathrm dx[/tex]
[tex]\implies f_Y(y)=\begin{cases}\dfrac6{(y+6)^2}&\text{for }y>0\\0&\text{otherwise}\end{cases}[/tex]
Then to get each conditional distribution, divide the joint distribution by the corresponding marginal distribution.
a.
[tex]f_{X|Y}(x,y)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}=\begin{cases}x(y+6)^2e^{-x(y+6)}&\text{for }x>0\\0&\text{otherwise}\end{cases}[/tex]
b.
[tex]f_{Y|X}(x,y)=\dfrac{f_{X,Y}(x,y)}{f_X(x)}=\begin{cases}xe^{-xy}&\text{for }y>0\\0&\text{otherwise}\end{cases}[/tex]
Part(a): The required answer is [tex]f_{x|y}(x,y)=(y+6)^2 xe^{-x(y+6)}[/tex]
Part(b): The required answer is [tex]f_{y|x}(x,y)=xe^{-xy}[/tex]
Given function is,
[tex]f(x,y)=6xe^-x(y+6)[/tex]
Part(a):
For finding the required value we will use the formula,
[tex]f_{x|y}(x,y)=\frac{f(x,y)}{f_y(y)}[/tex]
Now, solving above we get,
[tex]f_{x|y}(x,y)=\frac{f(x,y)}{f_y(y)}\\=\frac{6xe^{-x(y+6)}}{\int_{0}^{\infty }f(x,y)dx}\\=\frac{6xe^{-x(y+6)}}{\frac{6}{(y+6)^2}}\\=(y+6)^2 xe^{-x(y+6)}[/tex]
Part(b):
For finding the required value we will use the formula,
[tex]f_{y|x}(x,y)=\frac{f(x,y)}{f_x(x)}[/tex]
Now, solving above we get,
[tex]\\=\frac{6xe^{-x(y+6)}}{\int_{0}^{\infty }6xe^{-x(y+6)}dy}\\=xe^{-xy}[/tex]
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