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X and Y are two points that are selected according to a uniform distribution. Calculate the expected value of the distance between these two points for each of the following cases: a) Both X and Y are selected independently from the range 0-1. b) We select X between 0 and 1 and then select Y between 0 and X. Note: For case b the distance is equal to Y-X|

User DominiCane
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Answer:

the expected value of the distance between X and Y in this case is 0.

Explanation:

(a) When both X and Y are selected independently from the range 0-1, the distance between the two points can be calculated as the absolute difference between X and Y, denoted as |Y - X|. To find the expected value of this distance, we need to integrate over all possible values of X and Y and weight them by their respective probabilities.

The probability density function (PDF) for a uniform distribution on the interval 0-1 is given by f(x) = 1 for 0 ≤ x ≤ 1, and 0 otherwise.

Therefore, the expected value of the distance between X and Y in this case can be calculated as:

E(|Y - X|) = ∫∫ |y - x| f(x) f(y) dx dy

= ∫∫ |y - x| dx dy

= ∫ [∫ |y - x| dx] dy

Evaluating the inner integral:

∫ |y - x| dx = (x - y) / 2 | 0 ≤ x ≤ y + (y - x) / 2 | y ≤ x ≤ 1

= (y - y) / 2 + (1 - y) / 2

= (1 - y) / 2

Substituting this result back into the outer integral:

E(|Y - X|) = ∫ (1 - y) / 2 dy | 0 ≤ y ≤ 1

= [(1 - y)^2 / 4] | 0 ≤ y ≤ 1

= (1 - 1/4) - (1 - 1) / 4

= 3/8

Therefore, the expected value of the distance between X and Y when both are selected independently from the range 0-1 is 3/8.

(b) In this case, we select X between 0 and 1 and then select Y between 0 and X. The distance between the two points is given by Y - X.

To find the expected value, we need to integrate over the joint probability density function (PDF) of X and Y.

The joint PDF of X and Y can be expressed as:

f(x, y) = f(x) f(y|x) = 1 * 1/x = 1/x, for 0 ≤ x ≤ 1 and 0 ≤ y ≤ x

Now we can calculate the expected value of the distance:

E(Y - X) = ∫∫ (y - x) f(x, y) dx dy

= ∫∫ (y - x) (1/x) dx dy

Integrating the inner integral:

∫ (y - x) (1/x) dx = (y - x) ln(x) | 0 ≤ x ≤ y

= (y - y) ln(y) - (y - 0) ln(0) | 0 ≤ y ≤ 1

= 0

Therefore, the expected value of the distance between X and Y in this case is 0.

To summarize:

a) When X and Y are selected independently from the range 0-1, the expected value of the distance |Y - X| is 3/8.

b) When X is selected between 0 and 1, and Y is selected between 0 and X, the expected value of the distance Y - X is 0.

User Hellojinjie
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