<Transformation Technique>
-One Variable (Discrete, Continuous)
<Transformation between Discrete functions >
<One to One Function - Injective >
As long as the relationship between the values of X and Y=g(x) is one to one, we can easily deal with the discrete case. For this case, probability remains unchanged.
<Surjective Function - not one-to-one>
You should be careful here because the probability changes here.
<Transformation between Continuous functions >
To assume that the probability given by Y=U(X) is differentiable and either increasing or decreasing for all values within the range of X so that the we can find all corresponding values of y when it comes to inverse function X= W(Y) .
-Several Variables
It is the case for the joint pdf of several variables are given, and by using this, we determine the distribution of Y=u(X1,X2), which is the function of X. So by using the transformation method, we can find the distribution of Y by the joint pdf of X1,X2. Firstly we can find the joint pdf of Y and (X1 or X2) and then find the pdf of Y by using integration in terms of another X.
<Several to one>
<Several to Several>
Here we can derive Joint PDF of Y1, Y2 from Joint PDF of X1, X2 by using Jacobian. Then you can find the marginal PDF of X1 and Y1 using summation in the case of discrete or integration in the case of continuous. First we express X1,X2 interms of Y1 and Y2 which are the functions of X and you can finally find the joint pdf of Y1,Y2 by multiplying Jacobian to the joint pdf of X1,X2 expressed in terms of Y1, Y2.
<Prove relationship chi-square distribution and Normal distribution Using Transformation Technique>
As you know that the squared standard normal distribution is chi square distribution with degree of freedom 1. With this you can also derive that the sum of n square of independent standard normal distribution is chi square distribution with n.
<Every pdf of cumulative distribution function of continuous random variable =Unif(0,1)>
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