This figure below describes the joint PDF of the random variables X and Y. These random variables take values in [0,2] and [0,1], respectively. At x<=1, the value of the joint PDF is 1/2. At 2>x>1 the value of the joint PDF is 3/2. What's the marginal PDF of x if 1<x<2?

To find the marginal PDF of X when 1 < X < 2, we need to integrate the joint PDF with respect to Y over the range of [0,1].

Since the joint PDF is given as follows:
1/2 for X <= 1 and 0 <= Y <= 1
3/2 for 1 < X < 2 and 0 <= Y <= 1

For 1 < X < 2, the marginal PDF of X is given by integrating the joint PDF over the range of Y from 0 to 1:

marginal PDF of X = ∫(joint PDF) dy (from 0 to 1)

= ∫(3/2) dy (from 0 to 1)
= (3/2) * (1 - 0)
= 3/2

Therefore, the marginal PDF of X when 1 < X < 2 is 3/2.

It has to be in terms of x

Apologies for the confusion.

To find the marginal PDF of X when 1 < X < 2, we need to integrate the joint PDF with respect to Y over the range of [0,1].

Since the joint PDF is given as follows:
1/2 for X <= 1 and 0 <= Y <= 1
3/2 for 1 < X < 2 and 0 <= Y <= 1

For 1 < X < 2, the marginal PDF of X is given by integrating the joint PDF over the range of Y from 0 to 1:

marginal PDF of X = ∫(joint PDF) dy (from 0 to 1)

= ∫(3/2) dy (from 0 to 1)
= (3/2) * (1 - 0)
= 3/2

Therefore, the marginal PDF of X when 1 < X < 2 is 3/2.

To find the marginal PDF of X if 1 < X < 2, we need to integrate the joint PDF with respect to Y over the range [0,1].

The joint PDF is given as follows:

For X ≤ 1, f(x, y) = 1/2
For 2 > X > 1, f(x, y) = 3/2

Since we are interested in the marginal PDF of X, we need to find the probability density function for X.

First, let's express the joint PDF as a piecewise function:

f(x, y) =
1/2 if 0 ≤ x ≤ 1, 0 ≤ y ≤ 1,
3/2 if 1 < x < 2, 0 ≤ y ≤ 1,
0 otherwise.

Since we are only interested in the range 1 < X < 2, we are concerned with the second part of the joint PDF equation:

f(x, y) = 3/2 if 1 < x < 2, 0 ≤ y ≤ 1.

To find the marginal PDF of X, we integrate this equation with respect to y over the range [0,1]:

∫[0,1] (3/2) dy

Performing the integration, we get:

3/2 * [0,1]
= 3/2.

Therefore, the marginal PDF of X if 1 < X < 2 is 3/2.