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If the joint distribution of $x$ and $y$ as follow $$f(x,y)=\left\{ \begin{align} & \frac{{{x}^{2}}+y}{4}\,\,\,\,,\,\,\,\,0<x<y<4 \\ & 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,\,\,\,\,\,\,o.w. \\ \end{align} \right.$$

How can I find the marginal distribution function of $x$?

I tried integrating the function with respect to $y$ using the boundaries $0$ and $2$ and my answer did not match the back of the book.

Then I tried integrating with the boundaries $0$ and $2-y$ and my answer was closer but still wrong.

Where am I making my mistake?

Behrouz Maleki
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SwabianOrtolan
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  • Here is http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference for displaying functions. – Arbuja Jun 09 '16 at 23:21

2 Answers2

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It is important to identify the region where the joint density function "lives."

It was specified that the joint density function of $X$ and $Y$ is $\frac{x^2+y}{4}$ where $0\lt x\lt y\lt 2$. We are to understand that it is $0$ elsewhere.

Draw the line $y=x$. The joint density function lives in the part of the $2\times 2$ square that is above the line $y=x$. This is the triangle with corners $(0,0)$, $(2,2)$, and $(0,2)$.

For the marginal density of $X$, we need to "integrate out" $y$. For any $x$ between $0$ and $2$, $y$ travels from $y=x$ to $y=2$. So the marginal density of $X$ is $$\int_x^2 \frac{x^2+y}{4}\,dy$$ on the interval $(0,2)$, and $0$ elsewhere.

André Nicolas
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$${{f}_{X}}(x)=\frac{1}{4}\int_{x}^{2}{({{x}^{2}}+y)dy=\frac{1}{4}}{{x}^{2}}y+\left. \frac{1}{8}{{y}^{2}} \right|_{y=x}^{y=2}=-\frac{1}{4}{{x}^{3}}+\frac{3}{8}{{x}^{2}}+\frac{1}{2}$$ and $${{f}_{Y}}(y)=\frac{1}{4}\int_{0}^{y}{({{x}^{2}}+y)dx=\frac{1}{12}}{{x}^{3}}+\left. \frac{1}{4}xy \right|_{x=0}^{x=y}=\frac{1}{12}{{y}^{3}}+\frac{1}{4}{{y}^{2}}$$

Behrouz Maleki
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