Q: You have a random number generator that creates random numbers exponentially between \([0,1]\). You draw from this generator and keep adding the result. What is the expected number of draws to get this sum to be greater than 1.

Fifty Challenging Problems in Probability with Solutions (Dover Books on Mathematics)

A: Before getting into the solution for this, I'll go over an established theorem of exponential distributions.

If there exists two random variables which follow a exponential distribution with parameters \(\lambda_1\) and \(\lambda_2\) then their sum is given by the convolution of the two probability density functions. This is shown as

$$P(z = X_1 + X_2) = f_{z}(z) = \sum_{x=0}^{z}f_{X_{1}}(x) f_{X_{2}}(z - x)$$

The probability density function of a distribution with rate parameter \(\lambda\) is given as

$$f(k,\lambda) = \frac{\lambda^{k}e^{-\lambda}}{k!}$$

Plugging this into the convolution formula gives us

$$ f_{z}(z) = \sum_{x=0}^{z}\frac{\lambda_{1}^{x}}{x!}e^{-\lambda_{1}}\times \frac{\lambda_{2}^{z-x}}{(z-x)!}e^{-\lambda_2} $$

With some rearrangement of the terms the above simplifies to (hint: multiply & divide by \(z!\) and use the binomial expansion of \( (\lambda_1+\lambda_2)^{z}\) )

$$ \frac{(\lambda_1 + \lambda_2)^{z}}{z!} $$

which is the same as Poisson(\(\lambda_1 + \lambda_2\)). Coming back to the problem, we want to find \(n\) such that \(\lambda_1 + \lambda_2 + \ldots + \lambda_n = 1\) and as we draw the sum just once we can set \(k= 1\) in the probability density equation. This results in a probability density function

$$ P(k=1,\lambda_1 + \lambda_2 + \ldots + \lambda_n=1) = \frac{1^1 e^{-1}}{1!} = \frac{1}{e}$$

Which in turn implies that the number of draws needed to get the sum to 1 is \(e\). This fundamental number surfaces again!

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numero=zeros(10000000,1);

ReplyDeletefor j=1:10000000

suma=0;

i=0;

while(suma<1)

i = i + 1;

suma = suma +rand;

end

numero(j)=i;

end

mean(numero)-exp(1)

¬ 1e-4

Thanks for fixing the type of distribution. Do you know what the result might be if it were indeed a uniform distribution, or is there any difference? I followed Enrique in testing this programatically and encountered the same result, but I thought that random number generators for most languages pick numbers uniformly. Admittedly I've made zero effort in answering my own questions, this was just a passing wonder.

ReplyDeleteIts the same if it were uniform. The simplex way is the popular "text-book" way out there. I think something can be done for the uniform distribution case by observing that the natural log of a uniform random variable is exponentially distributed. But I'ven't given it much thought yet.

ReplyDelete