Question

X has cumulative distribution function f(x) = 0 if x < 1,

5/3(1-1/x^2) if x element of [1,2],

1 if x >2.

a. find probability density function of X

b. find E[X]

Answer 1

To find the probability density function (pdf) for X, we differentiate the given cdf, and for the expected value E[X], we integrate the product of the pdf and x over the interval [1,2].

Step-by-step explanation:

The student is asking to find the probability density function (pdf) and the expected value (E[X]) for a given cumulative distribution function (cdf) of a random variable X. A cumulative distribution function provides the probability that a random variable is less than or equal to a certain value, whereas the probability density function is the derivative of the cdf for continuous random variables and represents the probability at a given point. To find the pdf from the cdf, we differentiate the cdf with respect to x. In this case, we first differentiate f(x) = 5/3(1-1/x^2) for x in the interval [1,2] to get the pdf. To find the expected value E[X], we integrate the pdf over all possible values of X times x, which, for this problem, is the interval [1,2].

The solution steps include:

1. Differentiate 5/3(1-1/x^2) with respect to x to find the pdf f(x).
2. Integrate the pdf f(x) multiplied by x over the interval [1,2] to find E[X].