Question

Probability Density Function:

The probability density function of a random variable is given by: f(x) = k for 0 ≤ x ≤ 1 and f(x) = 0.5x for 1 ≤ x ≤ 2. Calculate the value of k to ensure this function is a valid probability density function.

Answer 1

To calculate the value of k to ensure that the given probability density function (pdf) is valid, you need to ensure that the area under the pdf is equal to 1. By integrating the pdf over the intervals 0 ≤ x ≤ 1 and 1 ≤ x ≤ 2, we can solve for k. The value of k is found to be 0.25.

Step-by-step explanation:

To calculate the value of k to ensure that the given probability density function (pdf) is valid, we need to ensure that the area under the pdf is equal to 1. We have two intervals for x: 0 ≤ x ≤ 1 and 1 ≤ x ≤ 2. In the first interval, f(x) = k, and in the second interval, f(x) = 0.5x. To calculate the area under the pdf for each interval, we integrate the respective functions.

For the interval 0 ≤ x ≤ 1:

∫b f(x) dx = ∫01 k dx = kx ∫01 = k ∫01 = k(1-0) = k

For the interval 1 ≤ x ≤ 2:

∫b f(x) dx = ∫12 0.5x dx = [0.5(x^2/2)] ∫12 = [0.25(2^2) - 0.25(1^2)]

Since the total area under the pdf must be equal to 1, we set the sum of the two integrals equal to 1 and solve for k:

k + [0.25(2^2) - 0.25(1^2)] = 1

k + [0.25(4) - 0.25(1)] = 1

k + 1 - 0.25 = 1

k + 0.75 = 1

k = 0.25