Cumulative distribution problem Statistics

[stats_student]

Homework Statement

Let X be a continuous random variable with cumulative distribution function given by F(x) = P(X<x).
Define a new random variable U=F(X).

Homework Equations

The Attempt at a Solution

OK so to solve this problem I first say U=F(X).

F(u)=P(F(X)<u)

which means

F(u)=P(U<u)

is this how i should solve this problem?

 

[andrewkirk]

You haven't stated what question is being asked of you.

Also, you will need to use more careful notation to avoid getting confused between the cdf of X and the cdf of U.
I suggest using a subscript, so that $F_U$ is the cdf of U, ie $F_U(u)=Pr(U<u)$.

 

[stats_student]

OK so to solve this problem I first say U=F(X).

F(u)=P(U<u)

which means

F(u)=P(F(X)<u)

is this how i should solve this problem?

 

[stats_student]

ignore that top post mis clicked.

However the questions asks to find the supports of U and the median.

I'm a little confused how to approach but can i say that u<x?

If that's not the case can you point me in the right direction.

Thanks

 

[andrewkirk]

Well, start with the support. What are the maximum and minimum possible values for F(X)? The support of U will be the interval between those.

By the way, there's a big hint in the choice of the variable name U. Can you think of a simple distribution whose name starts with the letter U?

 

[stats_student]

the possible values for F(X) is negative infinity to infinity.

Uniform distribution?

 

[stats_student]

Also would i be able to say that U ranges from negative infinity to infinity also?

 

[andrewkirk]

the possible values for F(X) is negative infinity to infinity.

F(X) is a probability. What is the range of possible values for a probability?

Uniform distribution?

What is the range of possible values for a random variable whose distribution is the standard uniform distribution? How does that compare to the range of F(X)?

 

[stats_student]

oh i see, so F(X) the range would be 0 to 1.

the range of values for a random variable whose distribution is the standard uniform distribution is negative infinity to infinity.

not should what you mean when you say compare the two?

 

[andrewkirk]

The range of possible values for a standard uniform random variable is the real interval [0,1]. If you've never come across standard uniform random variables before then it won't help you to realize that U is one, so perhaps don't worry about that side of it. Go back to your equation:
$$F_U(u)=Pr(F_X(X)<u)$$

Now since $F_X$ is a monotonic increasing function, you can invert it. What happens if you apply ${F_X}^{-1}$ to both sides of the inequality inside the parentheses?

After that you'll have one more step, which involves using the fact that $Pr(X<x)=F_X(x)$.

 

[stats_student]

Fu(u)=Pr(F(U)<x)?

 

[andrewkirk]

No.

What is ${F_X}^{-1}\big(F_X(X)\big)$?

 

[stats_student]

X?

 

[stats_student]

inverse of u equals F(u)?

 

[stats_student]

ok so would it be correct to say that U=F(X)

then the support of this must be U[0,1]

and thus the median will be 1/2

and the 25th percentile would be 1/4 and the third percentile would be 3/4?

 

[Ray Vickson]

ok so would it be correct to say that U=F(X)

then the support of this must be U[0,1]

and thus the median will be 1/2

and the 25th percentile would be 1/4 and the third percentile would be 3/4?

You tell us.

 

[stats_student]

well because U is essentially the height of the function and F(X) can only be between 0 and 1. then it must be so that the height U must be equal to [0,1]? is that sound reasoning?

 

[stats_student]

anyone?

 

[stats_student]

or should i say that because F(U) ranges from 0 to 1 then u must range from 0 to 1?

 

[Ray Vickson]

anyone?

Always start with a picture; it will help you straighten out your thinking. Look at the correspondence between $X = x_0$ and $U = u_0 = F(x_0)$: