Download The joint pdf of pressures for right and left front t

Beautiful Homework #12
Problem 5.19
McIndoo
Page 1 out of 3
Problem 19
Problem statement: The joint pdf of pressures for right and left front tires is given in Exercise 9.
a. Determine the conditional pdf of Y given that X=x and the conditional pdf of X given that
Y=y.
b. If the pressure in the right tire is found to be 22 psi, what is the probability that the left tire
has a pressure of at least 25 psi? Compare this to P(Y≥25).
c. If the pressure in the right tire is found to be 22 psi, what is the expected pressure in the left
tire, and what is the standard deviation of pressure in this tire?
Once again, let us begin by first taking a look at the joint pdf.
0.016
0.014
0.012
0.01
f(x,y) 0.008
0.006
0.004
0.002
0
26
y
20 22
24
26
20
28
30
x
Figure 1-Joint pdf for 5.9 and 5.19
Part a
The conditional probability density function of an rv should seem very familiar. Recall from
section 2.4, page 69, that the definition of conditional probability for any two events A and B is
given by
P ( A ∩ B)
P( B)
The joint conditional pdf retains the same overall form but the notation is different because
instead of the probability of an event, we use the marginal pdfs of the random variables and the
joint pdf. The definition of conditional pdfs is given on page 207. So, in order to find the
conditional pdf of Y given that X=x, we use
P( A B) =
f Y X ( y x) =
f ( x, y)
f x ( x)
−∞< y< ∞
The conditional pdf of X given that Y=y is found in a similar manner. All that is required for
this particular problem is setting up the conditional pdf using the joint pdf from the problem
statement of 5.9 and the marginal distributions of X and Y determined in 5.9d and e. Therefore,
Beautiful Homework #12
f Y X ( y x) =
Problem 5.19
Page 2 out of 3
f ( x, y)
f x ( x)
 K ( x2 + y 2 )

= 10 Kx 2 + 0.05

0

f X Y ( x y) =
McIndoo
20 < y < 30
otherwise
f ( x, y)
f y ( y)
 K ( x2 + y 2 )

= 10 Ky 2 + 0.05

0

20 < x < 30
otherwise
Part b
We are asked to find the probability that the left tire has a pressure of at least 25 psi if the right
tire has a pressure of 22 psi. Once again, this wording should sound very familiar. To find the
probability, we set up the problem as follows:
K (( 22 2 ) + y 2 )
P(Y ≥ 25 X = 22) = ∫ f Y X =22 ( y ) dy = ∫
dy
2
25
25 10K ( 22 ) + 0.05
30
30
= 0.556
Next, we are asked to compare this value with that of finding just the probability that Y≥25.
P(Y ≥ 25) =
30
∫f
25
30
y
( y) dy = ∫ (10 Ky 2 + 0.05)dy
25
= 0.549
Clearly, there is some difference, although small, between these 2 probabilities. It is due to the
fact that X and Y are not independent, as determined in problem 9e. Therefore, from this result,
it makes sense to conclude that somehow the pressure in the right tire has an effect on the
pressure in the left tire.
Beautiful Homework #12
Problem 5.19
McIndoo
Page 3 out of 3
Part c
This problem also should feel familiar. We set it up y multiplying y by the conditional pdf at a
particular value of X and then integrate it over the given bounds. This process is nearly identical
to finding the expected value for a single continuous rv (page 150).
E (Y X = 22) =
∞
30
−∞
25
∫ y ⋅ fY X ( y 22) dy =
∫ y⋅
K (( 22 2 ) + y 2 )
10 K ( 22 2 ) + 0.05
= 25.37 psi
Next, we must find the standard deviation for the pressure in the right tire. The process is, once
again, nearly identical to finding the standard deviation of a single continuous rv. We will use
the shortcut method and thus we must first determine the value of E(Y2 ).
∞
K (( 22 2 ) + y 2 )
E (Y X = 22) = ∫ y ⋅ f Y X ( y 22) dy = ∫ y ⋅
10 K ( 22 2 ) + 0.05
−∞
20
2
30
2
2
V (Y X = 22) = E (Y 2 X = 22) − [E(Y X = 22)]
2
= 652.03 − 643.79
= 8.24 psi 2
σY
X =22
= V (Y X = 22)
= 2.87 psi
0.014
0.012
0.01
f(x,y)
0.008
0.006
0.004
0.002
0
σy=2.87
µ y=25.37
20
22
24
26
y
Figure 2-Problem 5.19c
28
30