Is it possible to obtain a pdf of a random variable Y expressed as:
Y=a*log(b+X),
where a,b are constants and X is a random variable with Chi-squared distribution.
Thanks
Yes, it possible via some transformation. The easiest is the Jacobi of transformation of X onto Y. Define f(x) and g(y) as the pdfs of x and y.
Set g(y)=f(x) |dx/dy|.
In few words, replace x in your chi-square distribution with (exp(y/a)-b) and multiple the result with 1/a exp(y/a). After some simplification, you will arrive at the desired distribution.
Best of luck
Yes, it possible via some transformation. The easiest is the Jacobi of transformation of X onto Y. Define f(x) and g(y) as the pdfs of x and y.
Set g(y)=f(x) |dx/dy|.
In few words, replace x in your chi-square distribution with (exp(y/a)-b) and multiple the result with 1/a exp(y/a). After some simplification, you will arrive at the desired distribution.
Best of luck
Firstly, define y=g(x)=a*log(x+b). Obviously we have the inverse function as x=g^(-1)(y)=exp(y/a)-b.
Secondly, the random variable X is chi-squared distributed. Denote its density by f(x) (you find this function in.wikipedia or any good book on statistics).
Next use the rule of "transformation of the probability density function" as described in
http://en.wikipedia.org/wiki/Probability_density_function
This is essentially the chain rule of differentiation. Your desired probability density will be
f(y) = f[g^(-1)(y)] * { g^(-1)(y) + b }/a
where the term in curly brackets is the first derivative of g^(-1)(y) with respect to y and for f( . ) you use the chi2-density with argument g^(-1)(y).
Inverse probability integral transform(inverse transform sampling) could be of help. Give it a try.
Thank you Markus and Smaila, your answers were very helpful. I was able to derive the final expression of the discussed pdf with the help of your suggestions and it looks it's correct, as I've verified it with an empirical pdf obtained by simulation.
Why not use the delta function method - it's far more direct - used often by physicists.
Write h(y) = Intrgral{ f(x) delta[y - g(x)] } dx
put g(x) = log(x) and re-write everything in f(x) in terms of x=exp(log(x))
then just pull out the resulting new form of f(y) inside the integral
The result s/b h(y) = 2**v-2/ gamma(v/2) * exp[ (-*v/2+1)y - 1/2 exp(-y) ]
This method works very nicely for g(x) = 1/x
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