Need help on a statistical algorithm.

[Bman28]

I'm not great on statistics so I thought I would ask here.

For a project, I need to create a random number generator code. But I need the mean to be docile.

For instance. A RandomNumberGenerator that generates a float (Decimal) value between 0 and 1. But the results are skewed in favour of (For example) 0.1 when the mean is calculated. But if I changed a variable I can manipulate the mean to be a different number.

If this is something challenging then don't worry. I was just wondering if there's a known algorithm already for this as I can't find it anywhere. But if there is one, Could you please help explain all the variables as, Like I said at the start, I'm really not great with statistics maths, especially the terminology .

Thanks for taking the time to read, And I apologise if this was the wrong forum to post this.

 

[Buckeye]

The way that I think about this is to draw random numbers from a beta distribution. First, we can solve for the parameters to give a specific mean using something similar to these approaches: https://stats.stackexchange.com/que...beta-distribution-using-the-mean-and-variance

And then use those parameters in this function: https://www.rdocumentation.org/packages/rBeta2009/versions/1.0/topics/rbeta

All of this assumes you have access to R.

 

[AngleWyrm]

double shapedRNG(Random rng, double alpha, double beta){
    double x = rng.nextDouble();
    return x^(alpha-1) * (1-x)^(beta-1);
}

 

[Dason]

double shapedRNG(Random rng, double alpha, double beta){
    double x = rng.nextDouble();
    return x^(alpha-1) * (1-x)^(beta-1);
}

That's just returning the unnormalized pdf at a uniformly random point. That won't generate random draws from the distribution itself though.

 

[Dason]

What programming language are you using? Are you allowed to use libraries or does it *all* have to be from scratch?

 

[Bman28]

What programming language are you using? Are you allowed to use libraries or does it *all* have to be from scratch?

Im using Lua, here are the math methods: https://www.tutorialspoint.com/lua/lua_math_library.htm

Again, Don't worry about making it yourself. I just need the formula similar to AngleWyrm. I've tried theirs and it isn't working. The mean is almost always extremely low or close to infinite.

 

[Dason]

Yeah AngleWyrm's code doesn't generate random samples from the beta distribution. It looks like Lua has some distributions programed in: https://scilua.org/sci_dist.html

Unfortunately the beta distribution isn't one that you can use the qrng on directly. If we could generate from the gamma distribution there is a fairly simple way to get draws from the beta as well... But it looks like the gamma distribution also won't allow qrng. If we had the inverse cdf (or even the cdf itself) it is pretty easy to do as well. But I don't see those available either.

Fortunately there are other methods. I would suggest implementing a rejection sampling scheme here since we have the pdf and can generate uniform random draws.

 

[Dason]

It looks like Lua has some distributions programed in: https://scilua.org/sci_dist.html

Unfortunately the beta distribution isn't one that you can use the qrng on directly.

Actually I might not have read closely enough. I think you can directly sample - you just need to provide a prng instead of a qrng. If that's the case it should be as easy as creating the distribution object and calling it's sample method with a prng of your choosing.

 

[AngleWyrm]

To change location and scale of any value x
newX = (oldX - shiftLocationX) / scaleX

But the results are skewed in favour of (For example) 0.1 when the mean is calculated. But if I changed a variable I can manipulate the mean to be a different number.

skew = (mean - median) / standardDeviation
mean = median + skew * standardDeviation

To generate a normal distribution from a random variable, take a bell curve and flip it on it's side.

 

[Dason]

@AngleWyrm ....

Are you trying to help? There might be a crumb of usefulness in that post but almost every contribution to this thread that you've made is a net detriment in my mind. You post some things that are kind of related but aren't really useful to what they're trying to do. Maybe try adding some explanation to how it directly relates to the problem at hand here.

And seriously what is up with that normal distribution flipped on it's side? What's the point of that?

 

[Dason]

@Bman28 let me know if my suggestions on how to do it in Lua help or not. I'm not sure if any of that requires any external packages or whatever because I don't use Lua but wanted to help.

 

[Dason]

@AngleWyrm were you perhaps trying to illustrate inverse transform sampling but got the details wrong?

https://en.m.wikipedia.org/wiki/Inverse_transform_sampling

 

[katxt]

The triangular is another simple function. Put b = 0 (bottom), t = 1 (top) and set M = mode
Function tri(b, M, t)
If (t - b) * Rnd < M - b Then tri = b + (M - b) * Rnd ^ 0.5
Else: tri = t - (t - M) * Rnd ^ 0.5
End If
End Function

 

[Dason]

Honestly that's probably a better suggestion for this. Going the full Beta route might be nicer if it's possible to code easily but the triangle distribution would probably work quite well here. I guess it all comes down to the specifics of what attributes are necessary from the distribution.

 

[katxt]

If you can mathematically define a shape you like for your distribution, it's easy to generate random numbers from it. It doesn't even have to have an area of 1. If the high point is M, then generate a pair of uniform numbers, x from 0 to 1, and y from 0 to M. If y is under your graph keep x. If not, try again with a new pair.

 

[Dason]

If you can mathematically define a shape you like for your distribution, it's easy to generate random numbers from it. It doesn't even have to have an area of 1. If the high point is M, then generate a pair of uniform numbers, x from 0 to 1, and y from 0 to M. If y is under your graph keep x. If not, try again with a new pair.

Indeed! I even suggested that algorithm in this post: http://www.talkstats.com/threads/need-help-on-a-statistical-algorithm.78447/post-233103

 

[spunky]

@AngleWyrm ....

Are you trying to help? There might be a crumb of usefulness in that post but almost every contribution to this thread that you've made is a net detriment in my mind. You post some things that are kind of related but aren't really useful to what they're trying to do. Maybe try adding some explanation to how it directly relates to the problem at hand here.

And seriously what is up with that normal distribution flipped on it's side? What's the point of that?

^^ OMG THIS^^ SO MUCH OF THIS.

 

[spunky]

The triangular is another simple function. Put b = 0 (bottom), t = 1 (top) and set M = mode
Function tri(b, M, t)
If (t - b) * Rnd < M - b Then tri = b + (M - b) * Rnd ^ 0.5
Else: tri = t - (t - M) * Rnd ^ 0.5
End If
End Function

Honestly that's probably a better suggestion for this. Going the full Beta route might be nicer if it's possible to code easily but the triangle distribution would probably work quite well here. I guess it all comes down to the specifics of what attributes are necessary from the distribution.

Isn't the convolution of of 2 iid uniforms the triangular (or tent) distribution? Even just having a simple uniform generator over [0,1] would get you the triangular distr through convolutions.

 

[Dason]

Isn't the convolution of of 2 iid uniforms the triangular (or tent) distribution? Even just having a simple uniform generator over [0,1] would get you the triangular distr through convolutions.

It is. But like you said it must be iid and this ends up with a symmetric distribution which i don't think quite fits the needs here. Plus is a convolution really simpler than an if/else?

 

[katxt]

Indeed! I even suggested that algorithm in this post: http://www.talkstats.com/threads/need-help-on-a-statistical-algorithm.78447/post-233103

You're absolutely right. You did and I missed it. Perhaps my explanation might remove a little of the mystique. Cheers.