Finding the pdf of a transformed univariate random variable

In summary, the process of finding the probability density function (pdf) of a transformed univariate random variable involves applying the transformation to the original variable and utilizing the change of variables technique. This typically includes determining the new variable’s range, calculating the Jacobian of the transformation, and using it to adjust the original pdf accordingly. The resulting expression represents the pdf of the transformed variable, allowing for the analysis of its statistical properties.
  • #1
Hamiltonian
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TL;DR Summary
Confused as to how to obtain the cdf of a transformed random variable.
1701139076070.png

The above theorem is trying to find the pdf of a transformed random variable, it attempts to do so by "first principles", starting by using the definition of cdf, I don't understand why they have a ##f_X(x)## in the integral wouldn't ##\int_{\{x:r(x)<y\}}r(X) dx## be the correct integral for the cdf of Y.
 
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  • #2
No. The way ##r(x)## gets into the equation is only in the limit of the integral, ##r(x) \lt y##. Suppose ##r(x) \lt y##. What is the probability (density) of that? It is the probability (density) of that associated ##x## value, which is ##f_X(x)##. So those are the probabilities that we want to total for the integral. (Notice that the integral is with respect to ##dx##, not ##dy##)

PS. I may have ignored the situation where multiple ##x## values give the same ##r(x)##. That still works out because the integral limit allows the associated ##x## densities to be summed
 
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  • #3
Suppose ##Y=f(X)## and we know the pdf of X. To get the pdf of Y, we can find the CDF of Y ##P(Y<y)## then differentiate it wrt to y to get the pdf.
$$P(Y<y)=P(f(X)<y)$$
And depending on X you have to do an appropriate manipulation to get the cdf of Y. Here's and example, deriving the pdf of a ##\chi^2## function with a deg of freedom of one. This variable here is just ##Z^2##, Z is a standard normal distribution. The pdf of Z is a bit complex but you can find it here.
https://www.thoughtco.com/normal-distribution-bell-curve-formula-3126278 lets call this function f(x).
And cdf of f is ##\int_{-\infty}^{x}f(x) dx## and called F(x). Let X be a standard normal variable, and ##\chi=X^2##. So
$$P(\chi<y)=P(-\sqrt{y}<X<\sqrt{y})$$
which is ##F(\sqrt{y})-F(-\sqrt{y}## differentiating wrt y
$$P(\chi=y)\frac{1}{2\sqrt{y}} f(\sqrt{y})+\frac{1}{2\sqrt{y}} f(\sqrt{y})=\frac{1}{\sqrt{y}} f(\sqrt{y})$$
 
  • #4
Apologies the website gave gives the formula for the general normal distribution for the standard normal, take ##\sigma=1,\mu=0## in the equation
 

FAQ: Finding the pdf of a transformed univariate random variable

What is the general method to find the pdf of a transformed univariate random variable?

The general method involves using the change of variables technique. If you have a random variable X with a known pdf f_X(x), and you define a new variable Y = g(X), then the pdf of Y, denoted as f_Y(y), can be found using the formula f_Y(y) = f_X(g^(-1)(y)) * |d/dy[g^(-1)(y)]|, where g^(-1) is the inverse function of g, and |d/dy[g^(-1)(y)]| is the absolute value of the derivative of the inverse function.

What is the role of the Jacobian in finding the pdf of a transformed variable?

The Jacobian determinant is crucial when transforming variables, especially in multivariate cases. For a univariate transformation, the Jacobian simplifies to the absolute value of the derivative of the inverse transformation function. It accounts for the change in scale introduced by the transformation, ensuring that the resulting pdf integrates to 1 over the transformed variable's domain.

How do you handle transformations that are not one-to-one?

For transformations that are not one-to-one, the range of the transformed variable Y may correspond to multiple intervals of the original variable X. In such cases, you need to sum the contributions from all intervals. Specifically, if Y = g(X) and the equation y = g(x) has multiple solutions x_i for a given y, then the pdf of Y is given by f_Y(y) = Σ f_X(x_i) * |d/dy[g^(-1)(y)]| for all i.

Can you provide an example of finding the pdf of a transformed variable?

Sure! Suppose X is a random variable uniformly distributed over [0,1], and you want to find the pdf of Y = X^2. The pdf of X is f_X(x) = 1 for 0 ≤ x ≤ 1. The transformation Y = X^2 has the inverse X = sqrt(Y). The derivative of the inverse is d/dy[sqrt(y)] = 1/(2sqrt(y)). Therefore, the pdf of Y is f_Y(y) = f_X(sqrt(y)) * |1/(2sqrt(y))|, which simplifies to f_Y(y) = 1/(2sqrt(y)) for 0 ≤ y ≤ 1.

What are common pitfalls to avoid when finding the pdf of a transformed variable?

Common pitfalls include neglecting the absolute value of the derivative in the Jacobian, not correctly identifying the range of the transformed variable, and failing to account for multiple intervals in non-one-to-one transformations. Additionally, one must ensure that the transformation

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