After evaluating the integral and applying the normalization condition, we find k = 3, making the probability density function f(x) = 3(1+x)^(-4) valid.
To find the constant 'k' that makes the function f(x) = k(1+x)^(-4) a valid probability density function (PDF), we need to satisfy the normalization condition. This condition requires the integral of the PDF over its entire range to be equal to 1:
The integral to solve is:
∫[0 to ∞] k(1+x)^(-4) dx = 1
Breaking down the integral:
∫[0 to ∞] k(1+x)^(-4) dx = k ∫[0 to ∞] (1+x)^(-4) dx
Now, applying the power rule of integration:
= k [(1+x)^(-3)/(-3)] evaluated from 0 to ∞
Evaluating the limits:
= k [0 - 1/(-3)]
Simplifying:
= k (0 + 1/3) = k/3
Setting this equal to 1 (normalization condition):
k/3 = 1
Solving for k:
k = 3
Therefore, the constant 'k' that makes the function f(x) = 3(1+x)^(-4) a valid probability density function is 3.
The question probable may be:
An insurance company's claims are modeled by a continuous, positive, random variable X, whose probability density function is proportional to (1+x) ^−4 , for 0<x<∞. Use this information to answer the following questions. (a) To solve this, we first need to understand the first sentence. Saying the "probability density is proportional" means there is a constant in front. In other words, the probability density function looks like f(x)={ k(1+x) ^−4 , 0<x<∞
0 otherwise
Find k to make f(x) a valid probability density function. You will need to solve an improper integral with either your calculator or computer.