Question

What is P(y_1 >y _2 ∣y_ 1 <2y_ 2 ) for independent exponentially distributed random variables

y_ 1 and y _2 each with mean 5?

Options:
A) 1/4
B) 1/3
C) 1/2
D) 2/3

Answer 1

The probability P(y_1 > y_2 | y_1 < 2y_2) for independent exponentially distributed random variables y_1 and y_2 can be calculated as 2/15.

Step-by-step explanation:

To find the conditional probability P(y1 > y2 | y1 < 2y2), we can start by finding the joint distribution of y1 and y2. Since y1 and y2 are independent exponentially distributed random variables, their joint distribution is given by:

P(y1 < 2y2 AND y1 > y2) = P(y1 < 2y2) = P(y1 < 2y2 ∩ y2 < y1)

Let's denote the variable Z = y1 - y2. Then,

y1 < 2y2 ∩ y2 < y1 ⟺ -y2 < Z < y1

Using this information, we can calculate:

P(-y2 < Z < y1) = ∫∫ e-y1 e-y2 dy1 dy2

By solving this integral, we get:

P(y1 < 2y2 AND y1 > y2) = ∫0∞ ∫0y1 e-y1 e-y2 dy2 dy1 = 2/3

Finally, we can find the conditional probability:

P(y1 > y2 | y1 < 2y2) = P(y1 > y2 AND y1 < 2y2) / P(y1 < 2y2) = (2/3) / (5e-1) = 2/15