Let Y= XZ+X, where X∼Uniform{1,2,3} and Z∼Normal(1,4) are independent. Find the value of fX∣Y=2 (2).
Answers
Step-by-step explanation:
6 Jointly continuous random variables
Again, we deviate from the order in the book for this chapter, so the subsec-
tions in this chapter do not correspond to those in the text.
6.1 Joint density functions
Recall that X is continuous if there is a function f(x) (the density) such that
P(X ≤ t) = Z t
−∞
fX(x) dx
We generalize this to two random variables.
Definition 1. Two random variables X and Y are jointly continuous if there
is a function fX,Y (x, y) on R
2
, called the joint probability density function,
such that
P(X ≤ s, Y ≤ t) = Z Z
x≤s,y≤t
fX,Y (x, y) dxdy
The integral is over {(x, y) : x ≤ s, y ≤ t}. We can also write the integral as
P(X ≤ s, Y ≤ t) = Z s
−∞ Z t
−∞
fX,Y (x, y) dy
dx
=
Z t
−∞ Z s
−∞
fX,Y (x, y) dx
dy
In order for a function f(x, y) to be a joint density it must satisfy
f(x, y) ≥ 0
Z ∞
−∞
Z ∞
−∞
f(x, y)dxdy = 1
Just as with one random variable, the joint density function contains all
the information about the underlying probability measure if we only look at
the random variables X and Y . In particular, we can compute the probability
of any event defined in terms of X and Y just using f(x, y).
Here are some events defined in terms of X and Y :
{X ≤ Y }, {X2 +Y
2 ≤ 1}, and {1 ≤ X ≤ 4, Y ≥ 0}. They can all be written
in the form {(X, Y ) ∈ A} for some subset A of R
2
.
1