Transformation of a random variable

[_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!

 

[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

 

[_joey]

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

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

 

[SW VandeCarr]

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

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

 

[_joey]

All I did was to give an example of linearizing the exponential function.

The example is irrelevant but thanks.

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 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.

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

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

 

[_joey]

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

 

[_joey]

I can't get the Latex to work properly even after resending, but I believe the correct CDF of V where I'm taking F(V)=F(t;B) is

F(t;B)=5+1-e^{(-t/B)} if t&gt;3
...=5 if 0&lt;t\leq 3

I'm sure someone will correct this if it's wrong. BTW, this is not a mapping of one PDF into another.

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

 

[SW VandeCarr]

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

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

 

[_joey]

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

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

 

[SW VandeCarr]

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

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

 

[_joey]

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

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.

 

[SW VandeCarr]

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.

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

 

[_joey]

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

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.

 

[SW VandeCarr]

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.

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.

 

[_joey]

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.

I appreciate your time.


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


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

Any comments and suggestions are very much appreciated

Thanks again!