# Suppose that the random variables X and Y have joint p.d.f.f(x,y)=\begin

Suppose that the random variables X and Y have joint p.d.f.
$$f(x,y)=\begin{cases}kx(x-y),0<x<2,-x<y<x\\0,\ \ \ \ elsewhere\end{cases}$$
Find the marginal p.d.f. of the two random variables.

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Nathalie Redfern

Given :
$$f(x,y)=\begin{cases}kx(x-y),0<x<2,-x<y<x\\0,\ \ \ \ elsewhere\end{cases}$$
To find marginal P.D.F of x
$$\displaystyle{{f}_{{{x}}}{\left({x}\right)}}={\frac{{{1}}}{{{8}}}}{\int_{{-{x}}}^{{{x}}}}{x}{\left({x}-{y}\right)}{\left.{d}{y}\right.}$$
$$\displaystyle{\frac{{{1}}}{{{8}}}}{\int_{{-{x}}}^{{{x}}}}{\left({x}^{{{2}}}-{x}{y}\right)}{\left.{d}{y}\right.}$$
$$\displaystyle={\frac{{{1}}}{{{8}}}}{{\left[{x}^{{{2}}}{y}-{\frac{{{x}{y}^{{{2}}}}}{{{2}}}}\right]}_{{-{x}}}^{{{x}}}}$$
$$\displaystyle={\frac{{{1}}}{{{8}}}}{\left[{x}^{{{3}}}-{\frac{{{x}^{{{3}}}}}{{{2}}}}+{x}^{{{3}}}+{\frac{{{x}^{{{3}}}}}{{{2}}}}\right]}$$
$$\displaystyle={\frac{{{1}}}{{{8}}}}{\left({2}{x}^{{{3}}}\right)}$$
$$\displaystyle={\frac{{{x}^{{{3}}}}}{{{4}}}}$$
To find marginal P.D.F of y
$$\displaystyle{{f}_{{{y}}}{\left({y}\right)}}={\frac{{{1}}}{{{8}}}}{\int_{{{0}}}^{{{2}}}}{x}{\left({x}-{y}\right)}{\left.{d}{x}\right.}$$
$$\displaystyle={\frac{{{1}}}{{{8}}}}{\int_{{{0}}}^{{{2}}}}{\left({x}^{{{2}}}-{x}{y}\right)}{\left.{d}{x}\right.}$$
$$\displaystyle={\frac{{{1}}}{{{8}}}}{{\left[{\frac{{{x}^{{{3}}}}}{{{3}}}}-{\frac{{{x}^{{{2}}}{y}}}{{{2}}}}\right]}_{{{0}}}^{{{2}}}}$$
$$\displaystyle={\frac{{{1}}}{{{8}}}}{\left[{\frac{{{8}}}{{{3}}}}-{\frac{{{4}{y}}}{{{2}}}}-{0}-{0}\right]}$$
$$\displaystyle={\frac{{{1}}}{{{8}}}}{\left({\frac{{{8}}}{{{3}}}}-{y}\right)}$$
$$\displaystyle={\frac{{{1}}}{{{3}}}}-{\frac{{{y}}}{{{8}}}}$$
$$\displaystyle={\frac{{{8}-{3}{y}}}{{{24}}}}$$