# Thread: Density function solution check

1. ## Density function solution check

Suppose Y has density function:

f(y) = ky(1-y) when 0<=y<=1
and
f(y)= 0 elsewhere

a)Find the value of k that makes f(y) a probability density function
b)Find P(0.4<=Y<1)
c)Find P(Y<=0.4|Y<=0.8)
d)Find P(Y<0.4|Y<=0.8)

sol.

for part a I found k to be 6 (by making the integral = 1 on the interval 0,1 and solving for k

for part b also used the integral between 0.4 and 1 and got 0.648

c) d) is where I have the problem. I suppose both must have the same answer.

Since this is conditional probability can we state that:

P(Y<=0.4|Y<=0.8) = [P(y<=0.4) AND P(y<=0.8)]/P(y<=0.8) = P(0.4<=Y<=0.8)/P(y<=0.8)

so I find using integral P(0.4<=Y<=0.8)=0.544
and P(y<=0.8) = P(0<=y<=0.8)=0.896

then we divide 0.544/0.896= 0.607
I feel that I am doing something wrong here, but can't see what.

2. Originally Posted by somestudent2
Suppose Y has density function:

f(y) = ky(1-y) when 0<=y<=1
and
f(y)= 0 elsewhere

[snip]
c)Find P(Y<=0.4|Y<=0.8)
d)Find P(Y<0.4|Y<=0.8)

[snip]

c) d) is where I have the problem. I suppose both must have the same answer.

Since this is conditional probability can we state that:

P(Y<=0.4|Y<=0.8) = [P(y<=0.4) AND P(y<=0.8)]/P(y<=0.8) = P(0.4<=Y<=0.8)/P(y<=0.8)

so I find using integral P(0.4<=Y<=0.8)=0.544
and P(y<=0.8) = P(0<=y<=0.8)=0.896

then we divide 0.544/0.896= 0.607
I feel that I am doing something wrong here, but can't see what.
The answer to (c) and (d) is

$\frac{\Pr(Y \leq 0.4 \text \, \text{and}\, Y \leq 0.8)}{\Pr(Y \leq 0.8)} = \frac{\Pr(Y \leq 0.4)}{\Pr(Y \leq 0.8)}$ $= \frac{\int_{0}^{0.4} ky(1-y) \, dy}{\int_{0}^{0.8} ky(1-y) \, dy} = \frac{\int_{0}^{0.4} y(1-y) \, dy}{\int_{0}^{0.8} y(1-y) \, dy} = .....$