Repeated normal distributions

Hello all,

I never was a huge fan when it came to chances, and I was never particularly good at it. Now, I actually need to do something but I can't figure out how.
I have a normal distribution, given the standard deviation and the mean. This represents the chance of a random number x being selected. This is mathematical, not computer science, so there are no limits on the number being selected or its precision. x is a real number. Let's say P(x) is the chance of number x being selected.

Now what I need to find out; given a numbers y and n, the chance that after selecting n numbers x, (x0, x1, ..., xn) with each a chance of being selected of P(x), these numbers multiplied (x0*x1*...*xn) result in the number y.

I've got some calculations on paper, but I think I'm not even on the right track... Anybody here has any idea on how to go about this?


Thanks

For some reason the mathematical constant 'e' springs to mind. That probably doesn't help much, but I'm sure there's a correlation there. :p

Good luck.

I'd like to extend EVOEx question and ask for any advice on a good book(s) on statistics that starts with the basic principles and goes on to more advanced topics. I too am particularly interested in studying this field.

Although this wouldn't be the only step, could you use Bayes Theorem to do part of this?

P(y | x) = (P(x | y) * P(y)) / P(x)

(x is bold signifying it is a vector)

Quote:

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Originally Posted by DavidP

Although this wouldn't be the only step, could you use Bayes Theorem to do part of this?

P(y | x) = (P(x | y) * P(y)) / P(x)

(x is bold signifying it is a vector)

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Hmmm I have to admit I have no idea on how to go about that.

Thanks for all help so far.

But I started with a simple case: two numbers, x0 and x1, where x0*x1 = y. I took the formula of a standard deviation (yes, Sebastiani; with e constant ;-), calling it p(x).
Then I did p(x1)*p(y / x1). Then if you take the sum of this result for every possible x1, that should be the answer if I'm not mistaken.
However, there are an infinite number of x1's. And I couldn't find out a way to calculate the integral of this formula...

Thanks

The normal distribution is described with the probability density function f(x).
If X is Random value with Normal distribution
f(x) = dP(X<x)/dx

If X1 and X2 are 2 independent Normal values
P(X1*X2<y) = P(X1<x)*P(X2<y/x|X1<x) - You need to Integrate the right part for x from -Inf to +Inf
at least for x>0, y>0 ( for negative values formula needs a little adaptation)

Now because X1and X2 independent
P(X2<y/x|X1<x) = P(X2<y/x)

f2(y) = dP(X1*X2<y)dy

after you take this differential you probably could express f2(y) through f(y)

Then you can express fN(x) through f(x) and fN-1(x)
At least I suppose so...

If I remember correctly, multiplying random variables corresponds to convolving their distribution functions. But I would have to look it up.

Mario, do you want books on probability or statistics? What we're doing right here is probability and most assuredly not statistics.

Quote:

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Originally Posted by EVOEx

x is a real number. Let's say P(x) is the chance of number x being selected.

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Then P(x) == 0. For a continuous distribution, the probability of any specific real number being drawn from that distribution is always zero. For continuous distributions you must always think in terms of intervals or, equivalently, differential probabilities.

The solution to your specific problem, which is, what is the probability of selecting a set of numbers x0,x1,...xn, the answer is simply that the probability is zero.

Quote:

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Originally Posted by brewbuck

Then P(x) == 0. For a continuous distribution, the probability of any specific real number being drawn from that distribution is always zero. For continuous distributions you must always think in terms of intervals or, equivalently, differential probabilities.

The solution to your specific problem, which is, what is the probability of selecting a set of numbers x0,x1,...xn, the answer is simply that the probability is zero.

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Thanks to everyone, again.

@brewbuck:
You're right. But, yes, it won't be continuous completely. But there will be a LOT of possible values. I know that wasn't what I said in my initial post, because I wanted to approach this in a proper way. But like you said, that was results in nothing useful.

@vart:
Okay, I'll have to look a bit closer to that to understand that post, it looked quite tough. I'll have to do that later. Thanks though!


Thanks everyone