Probability Density Functions

[Fuquan22]

Homework Statement

The number X is chosen at random between 0 and 1. Determine the probability density function of each of the random variables v=X/(1-X) and W=X(1-x).

Homework Equations

The Attempt at a Solution

The solution in the back of the book says "The random Variable V satisfies P(V< or = v) = P(X< or = v/(1+v)) = v/(1+v) for v > or = 0. Its density function is equal to 1/((1+v)^2) for v>0 and 0 otherwise. The random variable W satisfies P(W < or = w) = 1-sqrt(1-4w) for 0< or = w< or = (1/4) and its density function is equal to 2(1-4w)^(-1/2) for 0<w<(1/4) and 0 otherwise. Can someone walk me through the steps to get to this solution?

 

[mighty2000]

First, let's define the random variables v and W mathematically:

v = X / (1 - X)
W = X (1 - X)

To find the probability density function (PDF) of v, we need to find the probability that v takes on a certain value. This can be written as P(v < or = x) where x is the value that v takes on.

Using the definition of v, we can rewrite this as P(X / (1 - X) < or = x). This can be further simplified to P(X < or = x(1 + v)).

Since X is chosen at random between 0 and 1, we know that the probability of it being less than or equal to a certain value is just that value. So we can rewrite the above expression as x(1 + v) for 0 < or = x(1 + v) < or = 1.

Now, we can solve for v in terms of x and write it as v = x / (1 - x). This is the same as our original definition of v, so we can substitute it in to get P(v < or = x) = x / (1 + x) for 0 < or = x < or = 1.

This is the cumulative distribution function (CDF) of v. To find the PDF, we can take the derivative of this with respect to x. This gives us 1 / (1 + x)^2 for 0 < or = x < or = 1.

However, the probability of v taking on a negative value is 0, so the PDF is 0 for v < 0. Therefore, the final PDF of v is:

f(v) = 1 / (1 + v)^2 for v > 0 and 0 otherwise.

To find the CDF and PDF of W, we can use a similar approach. First, we write the CDF as P(W < or = w) = P(X(1 - X) < or = w). This can be rewritten as P(X < or = (1 + sqrt(1 - 4w)) / 2) since X is positive and less than 1.

We can now solve for w in terms of X and write it as w = (1 - X)^2 / 4. Substituting this in gives us P(W < or = w) = P(X < or