Transform Random Var CDF to Standard Normal: F(x)=1-exp(-sqrt x)

Click For Summary
SUMMARY

The discussion focuses on transforming the cumulative distribution function (CDF) of a random variable defined by F(x) = 1 - exp(-sqrt(x)) into a standard normal distribution, characterized by a mean of 0 and a standard deviation of 1. The transformation is expressed as y = g(x) = Φ⁻¹(F(x)), where Φ(y) represents the CDF of the standard normal distribution. It is established that while the transformation can be defined, the inverse function Φ⁻¹(1 - exp(-x)) cannot be simplified into standard functions, as proven mathematically.

PREREQUISITES
  • Understanding of cumulative distribution functions (CDFs)
  • Familiarity with the properties of the standard normal distribution
  • Knowledge of inverse functions and their applications in probability
  • Basic concepts of exponential functions and transformations
NEXT STEPS
  • Study the properties of the standard normal distribution and its CDF, Φ(y)
  • Learn about the mathematical proof regarding the non-expressibility of Φ⁻¹ in terms of standard functions
  • Explore transformations of random variables in probability theory
  • Investigate numerical methods for approximating Φ⁻¹ for practical applications
USEFUL FOR

Statisticians, data scientists, and students in probability theory who are interested in understanding transformations of random variables and their applications in statistical analysis.

rvkhatri
Messages
3
Reaction score
0
How to transform a random variable CDF to a standard normal
Given F(x) = 1- exp (-sqrt x), for x greater that 0

Thanks.
 
Physics news on Phys.org
rvkhatri said:
How to transform a random variable CDF to a standard normal
Given F(x) = 1- exp (-sqrt x), for x greater that 0

Thanks.

Welcome to MHB, rvkhatri! :)

What do you mean by "standard normal"?

Do you mean the PDF?
Or perhaps an equivalent normal distribution?
 
I like Serena said:
Welcome to MHB, rvkhatri! :)

What do you mean by "standard normal"?

Do you mean the PDF?
Or perhaps an equivalent normal distribution?

I meant standard normal distribution i.e. mean = 0, sigma = 1

My class notes say,
if F(x) = 1- exp (-x), there could be one-to-one transformation to a standard normal distribution. But I am not able to get a start on this.
 
rvkhatri said:
I meant standard normal distribution i.e. mean = 0, sigma = 1

My class notes say,
if F(x) = 1- exp (-x), there could be one-to-one transformation to a standard normal distribution. But I am not able to get a start on this.

Suppose the transformation is given by $y=g(x)$.
That is, if X is distributed according to your exponential F(X), then we will have g(X) ~ N(0,1).

Let $\Phi(y)$ be the CDF of the standard normal distribution.

Then the transformation $g$ needs to be such that the standard normal cumulative probability up to y must be the same as the exponential cumulative probability up to x.
As a formula:
$$\Phi(y) = F(x)$$

In other words:
$$y = g(x) = \Phi^{-1}(F(x))$$
 
I like Serena said:
Suppose the transformation is given by $y=g(x)$.
That is, if X is distributed according to your exponential F(X), then we will have g(X) ~ N(0,1).

Let $\Phi(y)$ be the CDF of the standard normal distribution.

Then the transformation $g$ needs to be such that the standard normal cumulative probability up to y must be the same as the exponential cumulative probability up to x.
As a formula:
$$\Phi(y) = F(x)$$

In other words:
$$y = g(x) = \Phi^{-1}(F(x))$$

My class note gives me exactly this formula for trasformation.

Now how do we get value of y in terms of x.
 
rvkhatri said:
My class note gives me exactly this formula for trasformation.

Now how do we get value of y in terms of x.

We already have.

You're probably thinking of rewriting it into an expression using only standard functions.
But I'm afraid we can't.
The function $\Phi(x)$ cannot be expressed as a finite combination of standard functions (this has been proven mathematically).
As a result $\Phi^{-1}(1-e^{-x})$ cannot be expressed in such a form.

The expression we have is as simple as it gets.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
4K
  • · Replies 6 ·
Replies
6
Views
5K
  • · Replies 10 ·
Replies
10
Views
6K
  • · Replies 42 ·
2
Replies
42
Views
5K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K