Let $f(x,y)=cx(1-y), 0

Question

I've done a few problems similar to this before, but they have always have something simple like $0 \lt x \lt \infty$ and $0 \lt y\lt \infty$, where choosing the bounds to integrate over is straightforward. I'm getting a bit tripped up conceptually on $0<x<2y<1$.

$f(x,y)=cx(1-y), 0<x<2y<1$, and $0$ otherwise. Find $c$.

I did:

$$1=\int_0^{\frac12}\int_0^{2y}cx(1-y)dxdy$$ $$1=\int_0^{\frac12}c(1-y)\left[\frac{x^2}{2}\right]_0^{2y}dy$$ $$1=c\int_0^{\frac12}2y^2-2y^3dy$$ $$1=c\left[\frac{2y^3}{3}-\frac{y^4}{2}\right]_0^{\frac12}$$ $$1=c\left[\frac{1}{12}-\frac{1}{32}\right]\to c=\frac{96}{5}$$

From here, in order to determine if they're independent we need to find the marginal pdfs and see if $f(x,y)=f_1(x)f_1(y)$. At this point I got some funky numbers and I am unsure how to verify my conclusion that they're not independent.

I tried:

$$f_1(x)=\frac{96}{5}\int_0^{\frac12}x(1-y)dy=\frac{36x}{5}$$ $$f_1(y)=\frac{96}{5}\int_0^{2y}x(1-y)dx=\frac{192y^2}{5}-\frac{192y^3}{5}=\frac{192y^2}{5}\left(1-y\right)$$

Then to determine independence: $$f(x,y)=f_1(x)f_1(y)$$ $$\frac{96}{5}x(1-y)=\left(\frac{36x}{5}\right)\left(\frac{192y^2}{5}\right)\left(1-y\right)$$ $$\frac{96}{5}x(1-y) \neq \frac{6912xy^2}{5}(1-y)$$

Therefore, not independent.

Any help on where I may have gone wrong would be greatly appreciated. The last part of the problem asks me to find $P(X+Y \lt \frac12)$ and this is where I am fairly sure I'm messing up what to integrate over. Help on how to set up these integrals and a brief explanation on why you chose the values you did would be great, not necessary to solve them.

Answer 1

We show that the random variables, which I will call $X$ and $Y$, are not independent, in a "computation-light" way.

Draw a picture of where the joint density function of $X$ and $Y$ "lives." To do this, draw the rectangle with corners $(0,0)$, $(1,0)$, $(1,1/2)$, and $(0,1/2)$. Draw the line $2y=x$. This divides the rectangle into two triangles. The density function lives in the triangle that is above the line $2y=x$.

It is clear that the probability that $9/10\lt X\lt 1$ is not equal to $0$. It is also clear that the probability that $0\lt Y\lt 1/10$ is not equal to $0$.

However, the probability that $9/10\lt X\lt 1$ and $0\lt Y\lt 1/10$ is $0$, contradicting independence.

Much more informally, if we know that $X$ is "big" (bigger than $9/10$), then we know that $Y$ can't be small (less than $1/10$).

Answer 2

The marginal distribution of $x$ needs to be calculated by $f_1(x) = c\int_{x/2}^{1/2}x(1-y)dy$

If you plot the area where the joint distribution is defined, you will see that it is a triangle with vertices $(0,0), (0,1/2), (1,1/2)$. To find the marginal distribution of $x$, you need to integrate the joint distribution over the entire $y$ range and that is between the lines $y = x/2$ and $y = 1/2$.

To solve the last question plot the curve $x + y = 1/2$ onto the $xy-$plane you plotted for the first question. There you will see that you need to integrate the joint distribution over the area defined by $ x/2 < y < - x + 1/2$ and $0 < x < 1/3$.