Cumulative Standard Normal Distribution

[G Gabriel]

Re: Cumulative Standard Normal Distribution

Yeh..that's what happens when you are in a rush..Here is the corrected one:

FUNCTION Factorial AS N (value AS N )
dim x as n=1
dim k as n=max(value,1)
while x<k
	value=value*x
	x=x+1
	end while
	Factorial=value
END FUNCTION

[G Gabriel]

Re: Cumulative Standard Normal Distribution

Steve:
I can't even read the super-script in Excel.
What equation are you using for ERF()?

[Steve Andrews]

Re: Cumulative Standard Normal Distribution

Steve:
I can't even read the super-script in Excel.
What equation are you using for ERF()?

For Excel, there is a built-in function called ERF(). Just put the expression "+ERF(.50)" into a cell and you should see its value.

As for Xbasic, I supplied code above which was an attempt at creating an ERF() function. As stated, it isn't 100% accurate as of now but is a starting point for creating the function. I will attempt to refine it this evening.

[G Gabriel]

Re: Cumulative Standard Normal Distribution

Yes I used the ERF() in Excel but I am trying to create it in alpha and couldn't figure the equation used in Excel which I am assuming you are basing your function on. What is the equation?

[Steve Andrews]

Re: Cumulative Standard Normal Distribution

What is the equation?

There is no equation. If you want an expression for ERF(), forget it. As I said in my earlier post, it involves an "integration" for which there are no simple expressions. Using the Taylor series (a summation of expressions which converge onto the desired result) is about the only way.

I saw it mentioned somewhere that some people have equations which approximate the results but I have no idea on how accurate the approximations may be. To utilize the Taylor series within a loop seems to be the most accurate and cleanest approach. As I said, this evening I will try to get this function working.

[G Gabriel]

Re: Cumulative Standard Normal Distribution

This is the equation Excel uses:
http://upload.wikimedia.org/math/6/c...fb65ff7e6d.png
Gotta decipher it.

[Steve Andrews]

Re: Cumulative Standard Normal Distribution

Again, that equation involves an integration (the slender "S" symbol) for which there is no simple expression.

But I corrected my function! I had one sign incorrect. Try this out:

FUNCTION erf as N (X as N)
	dim nRet as N
	dim nTerm as N
	dim nConst as N
	dim nFactorial as N
	
	nRet = 0
	nTerm = 1
	nFactorial = 1
	nConst = 2/sqrt(math_pi)
	j = 0
	
	while abs(nTerm)>.000001 .AND. j < 20
		if j > 0
			nFactorial = nFactorial * j
		end if
		'nDenom = nFactorial*(2*j+1)
		nTerm = ((-1)^j)*(X^(2*j+1))/(nFactorial*(2*j+1))
		
		nRet = nRet + nTerm
		j = j + 1
	wend
	
	erf = nConst * nRet
END FUNCTION

Maybe Tom should put this one in his puzzle forum.

[Ted Giles]

Re: Cumulative Standard Normal Distribution

Now I know what you are up to, when you break the Las Vegas bank, can I have small �million?

[G Gabriel]

Re: Cumulative Standard Normal Distribution

Thanks Steve, I am not sure if Excel is using Taylor's Series, Simpson's Rule or who knows what. I will try your function. Yes it is a given that the answer in any scenario will be an approximation since you are measuring the area under the curve with infinite numbers. Just wanted to know what Excel does behind the scene and emulate that in alpha.

[G Gabriel]

Re: Cumulative Standard Normal Distribution

Steve
Just tried your function:

?erf(5)
= -237806564.970755 'should be 1

?erf(3)
= 0.708190838398696 'should be 0.99998

[Steve Andrews]

Re: Cumulative Standard Normal Distribution

Try commenting out the portion of the while statement limiting j to 20 iterations:

while abs(nTerm)>.000001 '.AND. j < 20

When I tested the function originally, I was passing parameters under 1.0. They converge rather quickly. But the examples you gave were wild - going from extremely high to extremely low. Thus, it took far more than 20 iterations to converge. For safety, you could still leave it as is but change the 20 to, say, 1000.

FUNCTION erf AS N (X AS N )
'DESCRIPTION: This function returns the Gauss Error Function of a number.
	dim nRet as N
	dim nTerm as N
	dim nConst as N
	dim nFactorial as N
	dim j as N
	
	nRet = 0
	nTerm = 1
	nFactorial = 1
	nConst = 2/sqrt(math_pi)
	j = 0
	
	while abs(nTerm)>.000001 .AND. j < 1000
		if j > 0	'When j=0, we should leave nFactorial=1.  
			nFactorial = nFactorial * j
		end if
		'nDenom = nFactorial*(2*j+1)		'For debugging purposes.
		nTerm = ((-1)^j)*(X^(2*j+1))/(nFactorial*(2*j+1))
		
		nRet = nRet + nTerm
		j = j + 1
	wend
	
	erf = nConst * nRet
END FUNCTION

[G Gabriel]

Re: Cumulative Standard Normal Distribution

Going to 1000 brings it very close.
You might want to tweak the function a little bit:
J =0 only once, in the first loop. There is no need to test the if statement in each loop thereafter.
I'd calculate the first term before the loop, then make j=1 and start the loop.
Also, you might want to terminate the loop once the term gets to zero so you don't have to run it a 1000 time.
Thanks

[G Gabriel]

Re: Cumulative Standard Normal Distribution

Also, you might want to terminate the loop once the term gets to zero so you don't have to run it a 1000 time

I see you are already doing that.

[G Gabriel]

Re: Cumulative Standard Normal Distribution

Here is a simplified version:

FUNCTION ERF as N (X as N)
	dim nRet as N=X
	dim nTerm as N=1
	dim nFactorial as N=1
	dim j as n=1

while abs(nTerm)>.000001 .AND. j < 200
	nFactorial = nFactorial * j
	nTerm = ((-1)^j)*(X^(2*j+1))/(nFactorial*(2*j+1))
	nRet = nRet + nTerm
	j = j + 1
	wend
erf = nRet*1.12838
END FUNCTION

[G Gabriel]

Re: Cumulative Standard Normal Distribution

And here is the function for CND:

FUNCTION NormS_Dist AS N (value AS N )
	NormS_Dist= 0.5 * (1 + erf(value/1.41421))
END FUNCTION

And if anyone cares, here is the one for Normal Distribution:

FUNCTION NormDist AS N (value AS N, mean AS N, SD AS N )
	NormDist=0.398942280401433/SD*2.718281828^((value-mean)^2*-1/2*SD^2)
END FUNCTION

And the one for Standard Normal Distribution:

FUNCTION STRD_NormDist AS N (value AS N)
	STRD_NormDist=0.398942280401433/1*2.718281828^(value^2*-1/2*1^2)
END FUNCTION