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

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Discussion Overview

The discussion revolves around transforming a cumulative distribution function (CDF) of a random variable, specifically F(x) = 1 - exp(-sqrt(x)), into a standard normal distribution. The scope includes theoretical aspects of probability distributions and transformations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant seeks guidance on transforming the CDF F(x) = 1 - exp(-sqrt(x)) to a standard normal distribution.
  • Another participant clarifies that "standard normal" refers to a distribution with mean 0 and standard deviation 1.
  • A participant mentions a transformation formula involving the inverse of the standard normal CDF, stating that if X follows the exponential distribution, then g(X) ~ N(0,1) can be expressed as y = g(x) = Φ^{-1}(F(x)).
  • It is noted that the transformation requires the standard normal cumulative probability to match the exponential cumulative probability.
  • One participant expresses a desire to express y in terms of x using standard functions but is informed that the inverse of the standard normal CDF cannot be expressed as a finite combination of standard functions.
  • Participants discuss the limitations of expressing the transformation in simpler terms, indicating that the current expression is as simplified as possible.

Areas of Agreement / Disagreement

Participants generally agree on the transformation approach but express uncertainty regarding the ability to simplify the expression further. There is no consensus on how to express y solely in terms of standard functions.

Contextual Notes

The discussion highlights limitations in expressing the inverse of the standard normal CDF and the transformation in simpler forms, which remains unresolved.

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

Thanks.
 
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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.
 

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