# Find the value of $c$ for the joint density function.

a) Find the value of $c$ for the joint density function.

$f(x,y)=\left\{ \begin{array}{lcc} c & if & x+y \leq 1, x,y\geq0 \\ \\ 0 & \text{other case} \end{array} \right.$

b) Find $P(X+Y\leq \frac{1}{2})$

My work

a) $f$ is a joint density function if $\int\int f(x,y)dxdy=1$ and $f(x,y)\geq 0$.

I need determine the area of integration. But i'm stuck here. because: $0\leq x+y\leq1$. Can someone help me with this?

Guide:

This is a picture of the described region, you just have to compute its area.

$$A=\frac12 bh$$

Edit:

Let $A$ be the area of triangle, $cA = 1$

• Uhmm, then $\int_0^1\int_0^1f(x,y)dxdy$? – Bvss12 Jul 13 '18 at 1:49
• You don't really need to use integration but if you want, $y$ takes value from $0$ to $1$, once we fix $y$, think of what are the values that $x$ can take in terms of $y$. The previous integration limits that you proposed describes a square of length $1$. – Siong Thye Goh Jul 13 '18 at 1:51
• I think, $c=1$ but, using the definition of density function, i think if $y$ takes the value $0$ to $1$ then $x$ takes the values $0$ to $1-y$ no? – Bvss12 Jul 13 '18 at 1:57
• $c$ is not equal to $1$ as the area of triangle is not $1$. your integration limits are correct now. $\int_0^1 \int_0^{1-y} f(x,y) \,\, dxdy$. – Siong Thye Goh Jul 13 '18 at 2:00
• yeah yeah sorry, is $c=2$ thanks for all (: – Bvss12 Jul 13 '18 at 2:02