Recent content by dabd

What is the fastest algorithm to find the closest root (such that the function value at that point is positive to an error but never negative, if not exactly zero) for a strictly decreasing function?

I was just interested in the calculation not in applying it to real data. The estimate is the median of the data x1,...,xn and I wanted to see how they derived that result.

How to compute \beta = arg min_\beta \sum_{i=1}^N {|y_i - x_i^T \beta|

I meant A^T A is symmetric.

I got the answer in simpler terms A^T A is symmetric, so it is positive semi-definite and by taking any \lambda > 0 the matrix A^T A + \lambda I is positive definite, hence non-singular and invertible.

That is not given. A is any n x m matrix.

if A is an n x m matrix where n < m I would like to prove that there exists some \lambda such that rank(A^T A + \lambda I) = m I know that if two of the columns of A^T A are linearly dependent, they are scalar multiples of each other and by adding some \lambda to two different positions, those...

By letting n=1 you eliminate the product and that is not what I meant. n is simply the upper limit of the product, i.e., 'i' goes from 1 to 'n'. Thanks.

Ok, so I am trying to integrate this with no success FullSimplify[ Integrate[ Product[PDF[NormalDistribution[y, \[Sigma]], Subscript[x, i]], {i, n}] PDF[NormalDistribution[\[Mu], \[Phi]], y]...

How to compute E[X|Y1,Y2]? Assume all random variables are discrete. I tried E[X|Y1,Y2] = \sum_x{x p(x|y1,y2) but I'm not sure how to compute p(x|y1,y2] = \frac{p(x \cap y1 \cap y2)}{p(y1 \cap y2)} If it is correct, how can I simplify the expression if Y1 and Y2 are iid?

If X is a random variable and f, g are functions is it possible to prove that: E[f(X)] > E[g(X)] \Rightarrow \exists x: f(x) > g(x)

I know this expression should return a Gaussian distribution but I can't get Mathematica to simplify the integral. What am I missing? \text{Simplify}\left[\frac{\text{Product}\left[\text{PDF}\left[\text{NormalDistribution}[y,\sigma...

Yes, the number is questionable, however even assuming 99,9% effectiveness - which I think is unrealistically optimistic, - the probability is at 0.63 which is still high, and this may be surprising for most people.

Yes, it is assuming the independence of the events. This is quite frightening news! It means that after a thousand protected encounters an individual almost surely has been infected, assuming the effectiveness of condoms is 98%. (Even if you raise it to 99% it doesn't affect the result...

I have read that the condom effectiveness in protecting from HIV infection is around 98%. Assuming the probability of contracting HIV from a single protected encounter is 2% the probability of getting nfected after 1000 protected encounters is (I took the math from here...