1. ## 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

2. 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.

3. 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??