Let Y possess a density function where f(y) = c(2-y) when y is between
0 and 2. and f(y) = 0 elsewhere.

a) Find c - I believe its 1/2 but i just want to double check
b) Find F(y) and use it to find P(1<=y<=2)

Thanks in advance!

Hi roadapples!!

First of all I will clarify the symbols that I will use:

int(a to b)[f(x)] means the integral from a to b of f(x).
[f(x)](b;a) means f(b)-f(a) 
+oo = positive infinite
-oo = negative infinite 


Now I can start:

f(y) is a density function if and only if it satisfies the following
requirements:

1) f(y) >= 0 

2) int(-oo to +oo)[f(y)] = 1

The condition 1) is obviously accomplished for y out of the interval
(0,2).

For 0 < y < 2, f(y) = c*(2-y) and (2-y)>= 0

then c must be c >= 0 .

To satisfy the contition 2):

1 = int(-oo to +oo)[f(y)] = int(0 to 2)[c*(2-y)] 
  = c*int(0 to 2)[(2-y)] =
  = c*(int(0 to 2)[(2)] - int(0 to 2)[(y)]) = 
  = c*([2*y](2;0) - 1/2[(y^2)](2;0)) =
  = c*((4-0) - 1/2*(4-0)) = 
  = c*(4-2) = 2*c

Then c = 1/2
This complete the part a) of your question.

By definition
F(y) = P(Y =< y)
F(y) = int(-oo to y)[f(x)] ;

In this case:

If y =< 0  then F(y)=0 ;

If 0 < y < 2 then 
F(y) = int(0 to y)[1/2*(2-x)] =
     = 1/2*([2*x](y;0) - [1/2*x^2](y;0)) =
     = 1/2*((2*y) - (1/2*y^2))=
     = y - 1/4*y^2 ;
F(y) = y - 1/4*y^2  if 0 < y < 2.

If y >= 2 then
F(y) = int(0 to y)[1/2*(2-x)] =
     = int(0 to 2)[1/2*(2-x)] + int(2 to y)[0] =
     = 2 - 1/4*2^2 = 2 - 1 = 1 ;
F(y) = 1  if y>= 2 . 

P(1<=Y<=2) = P(Y<=2) - P(Y<=1) =
           = F(2) - F(1) =
           = 1 - (1 - 1/4*1^2) =
           = 1 - (1- 1/4) = 1 - 3/4 =
           = 1/4 ;

P(1<=Y<=2) = 1/4 .

This complete the answer.

I hope this helps.
I did it based in my own knowledge, if you need a clarification, post
a request for it.

Best Regards.
livioflores-ga

Thats great stuff! Thanks!