Question

Consider f (x) = k (1 – x2 ) - 1 < x < 1.

0 Otherwise

Determine the value of k for which f (x) is a legitimate pdf.

Answer 1

To determine the value of k for which f(x) is a legitimate probability density function (pdf), we need to ensure that the function is non-negative for all values of x and the area under the curve is equal to 1. By evaluating the integral of the function over the given range and solving for k, we find that k = 3/5.

Step-by-step explanation:

To determine the value of k for which f(x) is a legitimate probability density function (pdf), we need to ensure that the following conditions are satisfied:

1. The function f(x) is non-negative for all values of x.

2. The area under the curve of f(x) is equal to 1.

For the given function, f(x) = k(1 - x^2), we only consider values of x in the range -1 < x < 1. For all other values of x, f(x) is equal to 0.

To find the value of k, we need to ensure that the area under the curve of f(x) is equal to 1. We can do this by evaluating the definite integral of f(x) over the given range:

1 = ∫ f(x) dx = ∫ k(1 - x^2) dx

Integrating the function,

1 = k * (∫ 1 dx - ∫ x^2 dx)

Using the integral rules,

1 = k * (x - (x^3)/3) | from -1 to 1

Substituting the limits of integration and simplifying,

1 = k * (1 - (1^3)/3 - (-1 - (-1^3)/3))

1 = k * (1 - 1/3 + 1/3)

1 = k * (1 + 2/3)

1 = k * (3/3 + 2/3)

1 = k * (5/3)

k = 3/5