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I have a random vector $(X,Y)$ with the density $f_{XY}(x,y)=\tfrac{1}{5}$ on the trapezoid $T_1$ with vertex $(0,0),(0,1),(2,1),(3,0)$ and $f_{XY}(x,y)=\tfrac{3}{2}x$ on the triangle $T_2$ with vertex $(0,4),(1,4),(1,3)$. I need to find:

  • support and probability function for marginal $X$;
  • support and probability function for $(Y|X=x), 0\le{x}\le1$;
  • support and distribution for $Z=X+Y$

I started with support for $T_1$ and $T_2$:

$T_1$ ${(x,y):0\le{x}\le{-y+3}, 0\le{y}\le{1}}$

$T_2$ ${(x,y):0\le{x}\le{1}, 3\le{y}\le{-x+4}}$

Now I need to check that $\iint_{T_1}f_{XY}(x,y)\mathrm{d}x\mathrm{d}y + \iint_{T_2}f_{XY}(x,y)\mathrm{d}x\mathrm{d}y = 1$, so:

$\frac{1}{5}\int_0^{1}(\int_0^{-y+3}{}\mathrm{d}x)\mathrm{d}y + \frac{3}{2}\int_0^1(\int_3^{-x+4}\mathrm{d}y)x\mathrm{d}x = 1$

For $T_1$ I got $\frac{1}{2}$ and for $T_2$ $\frac{1}{4}$, but the sum is not 1. Why?

I need to find the error before proceed with other questions :-)


First edit and first attempt:

  • support and probability function for marginal $X$:

$f_x(x)\begin{cases}\frac{3}{2}x^2+\frac{1}{5} & 0\leq{x}\leq{1}\\ \frac{1}{5} & 1\leq{x}\leq{2}\\ \frac{-x+3}{5} & 2\leq{x}\leq{3} \end{cases}$

  • distribution function for marginal $X$:

$F_x(x)\begin{cases}\frac{x^3}{2}+\frac{x}{5} & 0\leq{x}\leq{1}\\ \frac{7}{10} + \frac{x}{5} & 1\leq{x}\leq{2}\\ \frac{7}{10} + \frac{1}{5} -\frac{x^2}{10} + 3x & 2\leq{x}\leq{3} \end{cases}$

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    $\begingroup$ I think support for $T_2$ should be $0\leq x\leq 1$ and $-x+4\leq y\leq 4$. $\endgroup$
    – Mick A
    Commented Feb 18, 2016 at 22:38
  • $\begingroup$ Yes. Also the density is $\frac 32x$ $$\int_0^{1}\int_0^{-y+3}{\tfrac{1}{5}}\operatorname{d}x\operatorname{d}y + \int_0^1\int_{-x+4}^4 (\tfrac{3}{2}x)\operatorname{d}y\operatorname{d}x = \tfrac 1 2+\tfrac 1 2$$ $\endgroup$ Commented Feb 19, 2016 at 1:42
  • $\begingroup$ Hi, your suggestion was helpful, I edited my post. How do I proceed now? $\endgroup$
    – Paul
    Commented Feb 20, 2016 at 14:14

1 Answer 1

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  1. I think your marginal pdf for $X$ is right.

  2. I think your cdf for it is wrong, although the question doesn't require it. It should be:

$$ F_X(x) = \begin{cases} \int_{0}^{x} (3u^2/2+ 1/5)\;du &= \dfrac{x^3}{2} + \dfrac{x}{5}, & \text{if $0\leq x\leq 1$} \\ \\ F(1) + \int_{1}^{x} (1/5)\;du &= \dfrac{x}{5} + \dfrac{1}{2}, & \text{if $1\leq x\leq 2$} \\ \\ F(2) + \int_{2}^{x} (\frac{-u+3}{5})\;du &= \dfrac{-x^2}{10} + \dfrac{3x}{5} + \dfrac{1}{10}, & \text{if $2\leq x\leq 3.$} \\ \end{cases}$$

$$\\$$

  1. Conditional pdf of $Y$ given $X$ with $0\leq X\leq 1$:

In this range for $X$ we have two ranges for $Y$ to handle.

For $0\leq y\leq 1$:

\begin{align} f_{Y|X}(y|x) &= \dfrac{f_{X,Y}(x,y)}{f_X(x)} = \dfrac{1/5}{1/5 + 3x^2/2} = \dfrac{2}{2 + 15x^2}. \end{align}

For $4-x\leq y\leq 4$:

\begin{align} f_{Y|X}(y|x) &= \dfrac{f_{X,Y}(x,y)}{f_X(x)} = \dfrac{3x/2}{1/5 + 3x^2/2} = \dfrac{15x}{2 + 15x^2}. \end{align}

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  1. Support and pdf of $Z=X+Y$:

Assuming you've already drawn the regions above on the $xy-$plane, now imagine a line $x+y=c$ moving across. Values of $c$ are in the support of $Z$ if the line crosses a region of support for $(X,Y)$. So the support for $Z$ is $0\leq Z\leq 3$ and $4\leq Z\leq 5$.

It also shows the cases we should consider:

For $0\leq z\leq 1$:

$$P(Z=z) = \int_{x=0}^{z} P(X=x\cap Y=z-x)\;dx = \int_{0}^{z}\dfrac{1}{5}\;dx = \dfrac{z}{5}.$$

For $1\leq z\leq 3$:

$$P(Z=z) = \int_{x=z-1}^{z} P(X=x\cap Y=z-x)\;dx = \int_{z-1}^{z}\dfrac{1}{5}\;dx = \dfrac{1}{5}.$$

For $4\leq z\leq 5$:

$$P(Z=z) = \int_{x=z-4}^{1} P(X=x\cap Y=z-x)\;dx = \int_{z-4}^{1}\dfrac{3x}{2}\;dx = \dfrac{3}{4}\left(-z^2+8z-15\right).$$

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    $\begingroup$ Hi, finally I understand how this works, thank you very much. I think there is an error on CDF $F_x(x)$ if $1\leq x\leq 2$, it should be $F_X(x)=\frac{x}{5}+\frac{1}{2}$. I draw the CDF with R i.sstatic.net/NW1Bt.jpg $\endgroup$
    – Paul
    Commented Feb 21, 2016 at 11:35
  • $\begingroup$ @Paul You're right. Thanks for letting me know. I've made the change. $\endgroup$
    – Mick A
    Commented Feb 21, 2016 at 11:39
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    $\begingroup$ I verified that $f_Z(z)$ is a density function: $\int_{0}^{1}\frac{z}{5}\;dz+\int_{1}^{3}\frac{1}{5}\;dz+\int_{4}^{5}\frac{3}{4}(-z^2+8z-15)\;dz = \frac{1}{10}+\frac{2}{5}+\frac{1}{2} = 1$ $\endgroup$
    – Paul
    Commented Feb 21, 2016 at 14:37

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