Marginal Density and Conditional Density Problem

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Suppose (X, Y) is uniformly distributed over the region defined by $0 \leq y \leq \sqrt{1-x^2}$ and $-1 \leq x \leq 1$.

a. Find the marginal densities of X and Y

b. Find the conditional densities of $f_{X|Y = y}(x) and f_{Y|X = x}(y)$.

c. Are X and Y independent?

So I understand that the region is essentially a half circle and since it's a uniform distribution $f(x, y) = 1$ when $0 \leq y \leq \sqrt{1-x^2}$ and $0 \leq x \leq 1$ and $f(x, y) = 0$ otherwise.

For part A I'm a little confused on how to get the bounds for the integral to get the marginal densities, $f_X(x)$ and $f_Y(y)$. Would $f_X(x) = \int_0^\sqrt{1-x^2}dy$ and $f_Y(y) = \int_0^\sqrt{1-y^2}dx$?

For part B I know that $f_{X|Y=y}(x) = \frac{f(x,y)} {f_X(x)}$ and $f_{Y|X=x}(y) = \frac{f(x,y)} {f_Y(y)}$ If I am correct in part A, I believe $f_{X|Y=y}(x) = \frac{1}{\sqrt{1-y^2}}$ and $f_{Y|X=x}(y) = \frac{1}{\sqrt{1-x^2}}$ but I'm confused what the bounds are for this function to hold.

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but I'm confused what the bounds are for this function to hold.

The support, $\{\langle x,y\rangle: -1\leqslant x\leqslant 1~, 0\leqslant y\leqslant \surd(1-x^2)\}$, is the upper unit half-circle.

This may also be represented by: $\{\langle x,y\rangle: 0\leqslant y\leqslant 1~, -\surd(1-y^2)\leqslant x\leqslant \surd(1-y^2)\}$

[Notice: for any $y$, the values for $x$ range from negative to positive . ]


You have that $f_{\small X,Y}(x,y)= c\,\mathbf 1_{-1\leqslant x\leqslant 1, 0\leqslant y\leqslant\surd(1-x^2)}$ for some constant $c$ such that $\int_{-1}^1\int_0^{\surd(1-x^2)} c\,\mathrm d y\,\mathrm d x = 1$.

(Note: $c$ is the inverse of the support's area.)


You now have enough to correct your calculations.