I have the following joint PDF:
$$f_{UV}(u,v):= C \cdot \unicode[STIXGeneral]{x1D7D9}_{\{ (u,v): 0\le u \le 1, u \le v \le u+1 \}}(u,v)$$
(In case the subscript of the indicator function is a bit hard to read: $\{ (u,v): 0\le u \le 1, u \le v \le u+1 \}$)
I want to determine $C$ so this function can be considered a PDF. In other words, I need to show that:
$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x,y)dxdy=1$$
Here is my first problem:
I am having difficulties converting the subscript from the indicator function into an integral. My attempt:
$$C\int_0^1 \int_u^{u+1}1 \space dvdu=C\int_0^1 (v)\lvert_{u}^{u+1}du=C\int_0^11du=C$$
So in order for this to be normalized and satisfy the property, $C=1$. Is this correct or am I misreading the bounds from the indicator function?
Here is my second problem:
I want to determine the marginal PDF's $f_X$ and $f_Y$ from the joint PDF. I know this can be achieved by integrating over the "other" variable. For $f_Y$ the formula is:
$$\begin{align}%
\\
f_Y(y)=\int_{-\infty}^{\infty} f_{XY}(x,y)dx, \hspace{10pt} \textrm{ for all }y.
\end{align}$$
Similar problem to the first problem. My bounds are not independent from each other so I am not sure how to write the integral. How do I set up my bounds here? My attempt:
$$f_Y=\int_0^21dx=2 \\ \implies f_Y= \begin{cases}2 \hspace{10pt} \text{for} \hspace{10pt} 0 \le y\le2 \\ 0 \hspace{10pt} \text{else}\end{cases}$$
Is this correct?
Best Answer
The indicator is clear...what is wrong is that you have $f(u,v)$ and not $f(x,y)$
The support is a parallelogram (do a drawing to realize that) with area $1$ thus your density is uniform over this parallelogram and $C=1$, the reciprocal of the support area. No integrals are needed.
To get the two marginals you have to integrate
That is $U\sim U(0;1)$
For $V$ you have to observe that,
$$f_V(v)=\int_0^v du=v$$
$$f_V(v)=\int_{v-1}^1 du=2-v$$
concluding, $f_V$ is a triangular density
This is a compact way to write it
$$f_V(v)=[1-|1-v|]\mathbb{1}_{(0;2)}(v)$$