Distribution of sum of moments

[ll777]

Suppose there is a horizontal beam with length 1.

There is a pivot at distance $$\alpha$$ from the left end of the beam.

The beam is held in place, and n weights are placed on it. The positions of the weights (x₁,..,x_n) are independent and drawn uniformly from the interval [0, 1]. The beam is released, and either tips to the left or the right. I am interested in the probability that the beam tips to the left in terms of n and $$\alpha$$ when n is small.

Case A: Each weight has a mass of 1.

Here, I think the problem can be solved using the http://mathworld.wolfram.com/UniformSumDistribution.html" . When n=1, the beam tips left if the weight is left of the pivot. So, p(tips left)= p(x1<$$\alpha$$)=$$\alpha$$. When n=2, it tips left if x₁ + x₂ < $$\alpha$$. Let x₁ + x₂ = z. Since z is the sum of two uniform random variables, z has a triangular distribution, and this can be used to find p(tips left). Etc.

Case B: Everything is as above except that each weight to the left of the pivot is replaced with a weight of mass b > 1.

I want to find the probability the beams tips left in terms of n, $$\alpha$$ , and b. I am not sure of the best approach. One option is to repeatedly calculate convolutions, i.e. the distribution of the sum of moments for n + 1 is a convolution of the distributions for n and distribution of the moment of the (n+1)th weight, but I wonder if there is another way. Any suggestions would be much appreciated.

 

[bpet]

Case B: Everything is as above except that each weight to the left of the pivot is replaced with a weight of mass b > 1.

I want to find the probability the beams tips left in terms of n, $$\alpha$$ , and b. I am not sure of the best approach. One option is to repeatedly calculate convolutions, i.e. the distribution of the sum of moments for n + 1 is a convolution of the distributions for n and distribution of the moment of the (n+1)th weight, but I wonder if there is another way. Any suggestions would be much appreciated.

It could be done via conditional probabilities. Given that N points landed to the left of alpha, the combined moment about alpha could be written as M=-b*alpha*(U(1)+...+U(N))+(1-alpha)*(U(N+1)+...+U(n)) where U(1),...,U(n) are iid uniform.