Problem 3-5.1
ECE 3800
Western Michigan University
A random variable X has a probability density
function (pdf) of
𝑓𝑥 𝑥 = 2𝑥 0 ≤ 𝑥 ≤ 1
And a second independent variable Y is
uniformly distributed between -1.0 and 1.0.
a) Find the pdf of a random variable Z = X + 2Y.
b) Find the probability of 0<Z<1.
𝑤ℎ𝑒𝑛 𝑌 = 1 ⟶ 2𝑌 = 2
𝑌 = −1 ⟶ 2𝑌 = −2
0.6
0.5
0.4
fy(y)
Before we can
tackle the
summation of
the two
densities, we
have the deal
with the
Z = X + 2Y.
This is done by
modifying the
pdf for Y.
0.3
0.2
0.1
0
-3
-2
-1
0
1
2
y
By changing the range for the Y variable the
density function is stretched on the Y axis .
The Area of the rectangle must always equal 1
for the density function.
Now the span for Y is from -2 to 2, making the
base of the rectangle 4. the corresponding
height of the new rectangle is ¼.
3
Modified Density Function for y
0.6
0.5
fy(y)
0.4
0.3
0.2
0.1
0
-3
-2
-1
0
1
2
3
y
Density Function for x
2
1.8
1.6
1.4
1.2
fx(x)
With the
modified pdf for
Y we can now
convolve X and
Y.
We need to pick
one of the pdf’s
to flip. We’ll pick
fy(y).
1
0.8
0.6
0.4
0.2
0
0
0.5
1
x
1.5
2
We flip fy(y) and place it on the x-axis. Only x numbers can be on the x-axis. We need to reinterpret the yaxis into x values with x=z-y.
The leading edge of the pdf for Y becomes X=Z+2 and the trailing edge is X=Z-2.
Now we’re ready to slide the rectangle across the triangle.
2
1.8
1.6
1.4
fx(x)
fy(y)
1.2
1
0.8
0.6
0.4
0.2
0
Trailing edge
𝑦 = +2
𝑥 =𝑧−𝑦
𝑥 =𝑧−2
0
Leading edge
𝑦 = −2
𝑥 =𝑧−𝑦
𝑥 =𝑧+2
1
2
3
4
5
Region 1
The leading edge of the rectangle has broken into the triangle, but has not yet cleared x = 1.
2
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
-5.5
-4.5
-3.5
-2.5
-1.5
x
∞
𝑓𝑧 𝑧 =
𝑓𝑦 𝑧 − 𝑥 𝑓𝑥 𝑥 𝑑𝑥
−∞
𝑧+2
𝑓𝑧 𝑧 =
0
1
⋅ 2𝑥 ⋅ 𝑑𝑥
4
1
𝑓𝑧 𝑧 = 𝑥 2 |0 𝑧+2
4
1
𝑓𝑧 𝑧 = 𝑧 + 2
4
where − 2 ≤ 𝑧 < −1
2
where − 2 ≤ 𝑧 < −1
-0.5
𝑥=0
0.5
1.5
𝑥 =𝑧+2
Region 2
Once the leading edge of the of the rectangle break x=1 the area becomes constant until the trailing
edge slides past x=0. The trailing edge breaks into the stationary density function when (0=z-2) ,z=2.
2
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
-4.5
-3.5
-2.5
-1.5
-0.5
x
∞
𝑓𝑧 𝑧 =
𝑓𝑦 𝑧 − 𝑥 𝑓𝑥 𝑥 𝑑𝑥
−∞
1
𝑓𝑧 𝑧 =
0
1
⋅ 2𝑥 ⋅ 𝑑𝑥
4
where − 1 ≤ 𝑧 < 2
1
𝑓𝑧 𝑧 = 𝑥 2 |01
4
1
𝑓𝑧 𝑧 =
where 1 ≤ 𝑧 < 2
4
0.5
𝑥=0
1.5
𝑥=1
Region 3
Once the trailing edge slips past x = 0 the overlapping area decreases. This happens at
z = 2 and ends at z = 3, (1=z-2).
2
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
0.5
𝑥 =𝑧−2
1
1.5
𝑥=1
2
2.5
3
x
3.5
4
4.5
5
∞
𝑓𝑧 𝑧 =
𝑓𝑦 𝑧 − 𝑥 𝑓𝑥 𝑥 𝑑𝑥
−∞
1
𝑓𝑧 𝑧 =
1
⋅ 2𝑥 ⋅ 𝑑𝑥
𝑧−2 4
1
𝑓𝑧 𝑧 = 𝑥 2 |𝑧−21
4
1
𝑓𝑧 𝑧 = 1 − 𝑧 − 2 2
4
where 2 ≤ 𝑧 < 3
where 2 ≤ 𝑧 < 3
fz(z)
The pdf for fz(z) is made from several bounded functions
0.4
0.35
0.3
fz(z)
0.25
0.2
0.15
0.1
0.05
0
-3
-2
-1
0
1
z
1
𝑧 + 2 2 where − 2 ≤ 𝑧 < −1
4
1
𝑓𝑧 𝑧 =
where 1 ≤ 𝑧 < 2
4
1
𝑓𝑧 𝑧 = 1 − 𝑧 − 2 2
where 2 ≤ 𝑧 < 3
4
𝑓𝑧 𝑧 =
𝑓𝑧 𝑧 = 0
𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒
2
3
4