# Help with question~~

• Sep 10th 2010, 06:09 AM
freakmoister
Help with question~~
Hi everyone,

I need help with answering a question:

Question:

A pseudo-random number generates uniformly distributed random variables Xi on the interval [0,1]. This random number generator is to be used to produce random variables Yi, which have a Pareto distribution, described by

Probability density function:

p(y) = 0 where y < k
p(y) = ak^a/y^(a+1) where y >= k

Cumulative distribution function:

P(Y<=y) = 0 where y < k
P(Y<=y) = 1 - (k/y)^a where y >= k

with parameters a = 1.5 and k = 1.0
What value of Yi would correspond to an Xi value of 0.821? Explain your derivations.

==end of question==

Any help with this would be much appreciated.
Thanks
• Sep 10th 2010, 02:35 PM
mr fantastic
Quote:

Originally Posted by freakmoister
Hi everyone,

I need help with answering a question:

Question:

A pseudo-random number generates uniformly distributed random variables Xi on the interval [0,1]. This random number generator is to be used to produce random variables Yi, which have a Pareto distribution, described by

Probability density function:

p(y) = 0 where y < k
p(y) = ak^a/y^(a+1) where y >= k

Cumulative distribution function:

P(Y<=y) = 0 where y < k
P(Y<=y) = 1 - (k/y)^a where y >= k

with parameters a = 1.5 and k = 1.0
What value of Yi would correspond to an Xi value of 0.821? Explain your derivations.

==end of question==

Any help with this would be much appreciated.
Thanks

Use the probability integral transform theorem:

Suppose that Y is a continuous random variable with pdf f(y) and continuous cdf F(y). Suppose that X is a continuous standard uniform random variable. Then $V = F^{-1}(x)$ is a random variable with the same probability distribution as Y.

If you need more help, please show what you've tried and say where you get stuck.
• Sep 10th 2010, 07:45 PM
freakmoister
Hi,
thanks for the tip, but I'm not sure how to apply it in this case. Normally the cdfs/pdf would be y=f(x) but in this case the function has the variable y in it, so I'm not sure how to change the variables and do the inverse.

I tried just changing y to x such that I get the following

P(Y<=y) = 1 - (k/y)^a ---> P(X<=x) = 1 - (k/x)^a = 1 - (1/x)^1.5 (subbing in values)

so if y needs to be bigger than k to use the function then x must be similar therefore when Xi = 0.821 then the corresponding Yi = 0??? Dosen't look right because it seems too simple and I've not even used a lot of the given values.
When I sub in x=0.821 I get -0.344 which is not right.

I'm not really sure how to do the transformation. When you say V=F^(-1)(x) (i guess it refers to the cdf) does it mean I just replace all x within the function with y and replace y with x?

Say for example

Pr[Y<y] = F(x) = 2x-3 then F^(-1)(x) becomes ---> Pr[X<x] = F(y) = 2y-3
Then rearrange the function to be y = (x+3)/2

is this how it works??
What about the pdf?

I think I just have no idea what this inverse probability integral transform theorem is about.