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So I am given

$$ f_{X,Y}(x,y) =\begin{cases}\frac{1}{\pi}&\mathrm{\ if \ } x^2+y^2\le1\\ 0&\mathrm{\ otherwise\ }\end{cases} $$

And am asked to find joint probability density function for $X+Y$.

I'm assuming that I must use

$$ f_{X+Y}(z)=\int f_X(x)f_Y(z−x)\,\mathrm{d}x $$

however, I'm not sure where to go from there. I have seen examples of solving for $X+Y$ but only for examples when $X$ and $Y$ are independent and I can use the marginal densities. In this case would I solve for the marginal density of $X$ normally but for $Y$ solve in terms of $(z-x)$ rather than $y$?

wilsnunn
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Bacon
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    Wouldn't $\displaystyle f_{X+Y}^{,}(z)=\int f_{X,Y}^{,}(x,z−x),\mathrm{d}x$ be a more natural assumption? – Henry Dec 02 '15 at 00:29
  • I saw it written in both ways but wasn't sure which would be more useful. I guess I'm not sure what I'm integrating. If it's in regards to x^2+y^2≤1, would the limit be from 0 to √(1-y^2)? While integrating 1/π in respect to x? – Bacon Dec 02 '15 at 00:34
  • I believe I solved for the unit circle. But another part of the problem regards the unit sphere and gives the formulas for the volume of the interior in terms of V2k and V2k+1 (for even and odd respectively). It asks to solve for the joint probability density function of X1+...+Xn. How would the formula and process change in this case? – Bacon Dec 02 '15 at 05:33

2 Answers2

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Hint:

$\displaystyle f_{X+Y}^{\,}(z)=\int f_{X,Y}^{\,}(x,z−x)\,\mathrm{d}x$ is a more basic assumption.

You know that inside the disk the joint density is a uniform $\frac{1}{\pi}$, and that you are inside the disk, given $z$, when $x^2+(z-x)^2 \le 1$; this is a quadratic inequality which will give you an interval when $-\sqrt{2} \le z \le \sqrt{2}$, and so the integral is not much harder than solving quadratic equation.

Henry
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You need to solve: $$f_{X+Y}(z) = \int_{x^2+(z-x)^2\leq 1} f_{X,Y}(x, z-x)\operatorname d x$$

To find the integral bounds, solve $z^2-2xz+2x^2\leq 1$ for min/max $x$ where $z\in[-\sqrt 2;\sqrt 2]$

Graham Kemp
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