Exercise 2-4.2
A random variable has the probability density function:
( )
[ ( )
(
)]
0.3
0.25
0.2
0.15
0.1
0.05
0
-2
2
6
10
14
18
A second distribution can be described by:
Solve for the mean, mean-square, and variance of the distribution Y.
We will need to transform the density function fx into the density function fy.
( )
( )
For the transformation we need to invert the relationship between X and Y.
√
There are two parts to the transformation we need to deal with. The 1 st is the
derivative of X and 2nd is writing fx(x) in terms of Y.
𝑓𝑦 (𝑦)
𝑓𝑥 (𝑥 )
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑
√𝑌
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑓𝑥 (𝑥)
1
4
[𝑢(𝑥)
1 −1
𝑦 2
)], where 𝑋
𝑢(𝑥
𝑓𝑥 (𝑥)
𝑢
𝑦
√𝑌
𝑢
𝑦
Before we finish the density function for Y, we need to modify the
step functions. The step function triggers when the inside of the
parentheses becomes zero. Before this point the step is zero and
after this point the step is 1. Looking at:
𝑢
𝑦 and 𝑢
𝑦
𝑢
𝑦
𝑢
𝑦
𝑢(𝑦
6)
𝑢(𝑦)
In changing the steps they still trigger at the same values.
𝑓𝑦 (𝑦)
𝑓𝑥 (𝑥 )
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
[𝑢(𝑦)
𝑢 (𝑦
𝑓𝑦 (𝑦)
6)] ×
2
𝑦
𝑑𝑥 [𝑢(𝑦)
8𝑑𝑦
√𝑦
𝑑𝑥
𝑑𝑦
2
𝑢(𝑦
6)]
1.4
1.2
1
0.8
0.6
0.4
0.2
0
-2
2
6
10
14
18
Now we can finish writing the density function for Y.
To solve for the 1st Moment or mean of the function we start
with the formula:
̅
𝑌̅
( )
∫
16
∫ 𝑦
0
8 𝑦
𝑑𝑦
𝑌̅
16
8
1
∫ 𝑦 2 𝑑𝑦
0
𝑌̅
2
3
× × ( 6)2
8 3
Next we solve for the 2nd Moment or the Mean-Square Value
̅̅̅
∫
( )
5.333̅
̅̅̅
𝑌
16
∫ 𝑦
0
8 𝑦
𝑌̅
𝑑𝑦
16
8
3
∫ 𝑦 2 𝑑𝑦
0
𝑌̅
2
5
× × ( 6)2
8 5
Lastly we can now solve for the variance of the distribution
̅̅̅̅
5 .2
( ̅)
(5.333)
22.8
5 .2