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

AI Thread Summary
To transform the cumulative distribution function (CDF) F(x) = 1 - exp(-sqrt(x)) into a standard normal distribution, the transformation formula y = g(x) = Φ⁻¹(F(x)) is used, where Φ(y) represents the CDF of the standard normal distribution. The discussion clarifies that while the transformation can be defined, the inverse function Φ⁻¹ cannot be expressed using standard functions. Participants emphasize that the transformation is valid for x greater than 0. The conversation highlights the challenge of simplifying the expression further, as it has been mathematically proven that Φ cannot be expressed as a finite combination of standard functions. Understanding this transformation is crucial for applying it in statistical contexts.
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.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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