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I'm trying to find $f_Z(z)$ with $Z=2X-Y$, for $X$ and $Y$ with joint density function $f_{XY}(x,y)$: $$ \begin{cases} x/8 & 1 \le x \le 3 \land -1 \le y \le 1 \\ 0 & \text{elsewhere} \end{cases} $$

and I want to do it without using the convolution theorem.

My strategy was first finding the distribution of Z $F_Z(z)$ and then finding its derivative, which is the density function of Z $f_Z(z)$.

So, what I'm trying to find first is: $$F_Z(z)=P\{Z \le z\}=P\{2X+Y \le z\}=P\{(x,y) \in D_Z\} =\iint_{D_Z} f_{XY}(x,y) \,dx \, dy$$ with $D_Z$ being the half plane $2x-y \le z$.

I'm not sure how to find the limits of the integrals. I was trying something like this:

If $z \lt 1$, the joint density of $XY$ is zero, so the distribution of $Z$ is zero (I think I'm doing something wrong here). The plot is (sorry about the quality of the plots, I didn't knew how to make them look clearer): plot1

If $1 \le z \le 3$ the joint density of XY equals $x/8$. The plot is: plot2 In this case, $D_Z$ is the shaded area inside the rectangle. So, the distribution of Z for $1 \le z \le 3$, is: $$F_Z(z)=\int_{y=-1}^{y=1} \int_{x=1}^{x=\frac {z+y}{2}} \frac {x}{8} \, dx \, dy = \frac {z^2}{32}-\frac{5}{96}$$ Then, the density function of Z is $$f_Z(z)=\frac {z}{16}$$

But this result does not match the one that I obtained doing the same thing using the convolution theorem ($X$ and $Y$ are independent).

I also found the rest of the limits on $z$, but didn't do the integrals because I think that the only one that I did is wrong. They are: $$3 \lt z \le 5$$ $$5 \lt z \le 7$$ $$z \gt 7$$ All of these were found by finding the intersections of $z=2x-y$ with the corners of the rectangle (I can elaborate if necessary).

Thanks in advance to anyone that can be of help.

Michael Hardy
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totota
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2 Answers2

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If $1\le z\le3$ the limits on your double integral are wrong. It is true in the inner integral that $x$ runs from $1$ to $\frac{z+y}2$, but in the outer integral $y$ runs from $2-z$ to $1$. The lower limit for $y$ is $2-z$ because that's the value of $y$ when the line $2x-y=z$ intersects the line $x=1$. (The lower limit for $y$ is not $-1$ unless $z=3$.)

You have the right ranges on $z$ for the remaining cases.

grand_chat
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You have a number regions:

  1. $1\le z\le 3$:

$$ F_Z(z)=\int_{2-z}^1\int_1^{\frac{z+y}{2}}\frac{x}{8}dxdy=\frac{1}{192}z^3+\frac{1}{64}z^2-\frac{1}{64}z+\frac{5}{192} $$

  1. $3< z\le 5$:

$$ F_Z(z)=\int_{-1}^1\int_1^{\frac{z+y}{2}}\frac{x}{8}dxdy=\frac{1}{32}z^2-\frac{11}{96} $$

  1. $5< z\le 7$:

$$ F_Z(z)=\int_{-1}^{6-z}\int_1^{\frac{z+y}{2}}\frac{x}{8}dxdy+\int_{6-z}^1\frac{1}{2}dy=-\frac{1}{192}z^3+\frac{1}{64}z^2+\frac{35}{64}z-\frac{347}{192} $$

and $F_Z(z)=0$ for $z<1$ and $F_Z(z)=1$ for $z>7$.