# Transformation of a random variable

1. Sep 16, 2010

### _joey

The transformation of a random variable is well documented and there are numerous examples on the web. Most examples present univariate variable transformation utilising inverse of the transformation function. The method works whenever the transformation function is one-to-one.

Let's say g(x)=x^2 such that x>=0.

I am interested in solving a problem when the transformation function is not a monotone function ie not one-to-one on an interval.

For example, I have two random variables X and Y. X variable has a probability density function f(x)=0.2e^(-x/5) x>0, this is pdf of a classical exponential probability distribution. And a simple linear transformation function Y=3X for X>5 otherwise Y=7 when 0<X$$\leq5$$

Clearly, on [0, 5) g(.) transformation is not one-to-one. Would be possible to find pdf or cdf of variable Y only on (5,inf) or can it be determine over (0, infinity)?

Thanks!

2. Sep 16, 2010

### SW VandeCarr

Say:

$$y=ae^{kx}$$

Then:

$$ln(y)=ln(a)+ln(e^{kx})$$

$$ln(y)=ln(a)+kx$$

Let Y=ln(y) B=a

$$Y=B+kx$$

3. Sep 16, 2010

### _joey

thanks for you reply but I can't see how your equations are relevant. Could you please elaborate?

We have a transformation function g(.), in my case g(x)=k, that is not one-to-one and not probability density function. For a transformation to work one needs to have a monotone transformation function that maps one variable to another over specified interval.

Thanks again

4. Sep 17, 2010

### SW VandeCarr

All I did was to give an example of linearizing the exponential function. To convert it to a pdf you use $$\lambda e^{-\lambda x}$$. But, as you say, your function is not a pdf and not one to one, so the transformation won't work. I'm not sure what you're asking. You can define a function on the transformed expression any way you like as far as I can tell. Regarding pdfs, you usually want to transform them so as to preserve certain relationships, not introduce discontinuities and other distortions. In this case $$Y=\lambda-\lambda x$$

Last edited: Sep 17, 2010
5. Sep 17, 2010

### _joey

The example is irrelevant but thanks.

I know how to convert it using different methods but only when the transformation function is one-to-one. We have an intereval where transformation function g(x)=k is not one-to-one.

The question: what is the method of transforming a random variable over the interval where transformation function is not one-to-one. If not possible how to treat the discontinuity over entire interval preserving the relationship between two random variables.

PS Y=lambda-lambda*x is not a solution. Or a step to the right solution

Last edited: Sep 17, 2010
6. Sep 17, 2010

### _joey

Here's the question: http://img440.imageshack.us/img440/1879/cdf.gif [Broken]

Last edited by a moderator: May 4, 2017
7. Sep 17, 2010

### _joey

This is my solution for t>3. I can't figure out what to do with 0<t<=3 (Sorry, I can't use LaTeX on forum as I am very new to math forum discussions)

Solution for t>3
http://img823.imageshack.us/img823/7729/sol1.gif [Broken]

Last edited by a moderator: May 4, 2017
8. Sep 17, 2010

### SW VandeCarr

I just edited my post. Take another look for t less or equal to 3.

Last edited by a moderator: May 4, 2017
9. Sep 17, 2010

### _joey

I don't think either line is correct. In 0<t<=3 the probability of V=5 being occurred is integral of pdf from t=0 to t=3. The property of a probability density function is that it integrates to 1 over the interval it is defined. In my case it should be a sum over two intervals. Your cdf 5t/3 is not a cdf function over 0<t<=3

10. Sep 17, 2010

### SW VandeCarr

Why not? The distribution is uniform so the CDF 5t/3 is 0 when t=0 and 5 when t=3.

11. Sep 17, 2010

### _joey

The distribution is not uniform. The occurence of V=5 dependens on the probability of T occurring between 0<t<=3, and probability of T occurring on this interval has a pdf which is an exponential function. Moreover, F(.) is a cumulative function of probabilities and so 0<=F(.)<=1 for any value it takes. We are dealing with probabilities here.

12. Sep 17, 2010

### SW VandeCarr

Of course you're correct. If we just dealt with the uniform distribution portion the cumulative probability of t is t/3. However since it is part of a larger distribution we must consider the exponential portion as well and that the total integral must sum to one. So cumulative probability for t=0 to t <=3 is 1-F(v,B) where F(v,B) is the cumulative probability for t>3

13. Sep 17, 2010

### _joey

Dude, if you have anything worthwile to add then please do so, otherwise you are only re-iterating my comments. 5t/3 or whatever it is not a valid cdf function.

I am still trying to figure out what to do with the cdf over 0<t<=3 in transformation from one random variable to another.

14. Sep 17, 2010

### SW VandeCarr

Ok. Someone else may be able to help. Sorry I couldn't be of assistance. I know 5t/3 is not a valid cdf. I just made a mistake.

15. Sep 17, 2010

### _joey

I appreciate your time.

Here is the original question: http://img440.imageshack.us/img440/1879/cdf.gif [Broken]

My possible answer to the question above: http://img685.imageshack.us/img685/1043/sol2c.gif [Broken]

Any comments and suggestions are very much appreciated

Thanks again!

Last edited by a moderator: May 4, 2017