Question

The joint density function for a pair of random variables x and y is given f(x,y)= {cx(1+y) if 0≤x≤1, 0≤y≤2 0 otherwise find the value of the constant

c. find p(x+y≤1)

Answer 1

`f_(X,Y)(x,y)=\begin{cases}cx(1+y)&\text{for }0\le x\le1,0\le y\le2\\0&\text{otherwise}\end{cases}`


For
`f` to be a proper density function, we need to have the integral over its support
`\mathcal S` to equal 1.



`\displaystyle\iint_(\mathcal S)f_(X,Y)(x,y)\,\mathrm dx\,\mathrm dy=\int_(y=0)^(y=2)\int_(x=0)^(x=1)cx(1+y)\,\mathrm dx\,\mathrm dy=2c=1`



`\implies c=\frac12`


Now,



`\mathbb P(X+Y\le1)=\mathbb P(X\le1-Y)=\displaystyle\int_(y=0)^(y=2)\int_(x=0)^(x=1-y)\frac{x(1+y)}2\,\mathrm dx\,\mathrm dy=\frac13`