density function

[mark]

hello i was wondering if anyone could help me with problem

The problem says: Let Y possess a density function

f(y) = (c(2-y) for 0<y<2 )
0 o.w.

a. find c
b. find F(y)

i thought it was a beta distribution problem with alpha = 1, Beta = 2
but the beta distribution says that 0<x<1

 

[tkhunny]

Beta? Why does it have to have a name?

\(\displaystyle \L\;\int_{0}^{2}{c(2-y)}\;dy\;=\;1\)

Solve for 'c'. It's a constant.

 

[mark]

\(\displaystyle \L\;\int_{0}^{2}{c(2-y)}\;dy\;=1\)

1 = c(2y - (y^2)/2) evaluated from 0 to 2

1 = c(4-2)

c = 1/2

for part b i did

F(y) = \(\displaystyle \L\;\int_{-infinity}^{y}{f(t)}\;dt\;\)

for x <0

F(y) = 0

for 0<y<2

\(\displaystyle \L\;\int_{-infinity}^{y}{f(t)}\;dt\;\) = \(\displaystyle \L\;\int_{-infinity}^{0}{f(t)}\;dt\;\) + \(\displaystyle \L\;\int_{0}^{y}{f(t)}\;dt\;\)

= 0 + \(\displaystyle \L\;\int_{0}^{y}{(1/2)(2 - t)}\;dt\;\) = y - (y^2)/4

 

[tkhunny]

Did you define y > 2?

Nice work.

In this simple case, you can check your work with a little geometry. Remember the area of a trapezoid?