Cumulative probabilities/normal distribution/series of trials

[BvlyHls90210]

Assume a normal distribution. I understand how to calculate the probability of something being X standard deviations above the mean, for example, mean is 10, standard deviation is 5, probability of >15 is (1-.67)/2.

Furthermore, I understand that if we have say 100 trails, the new mean should be 100*the old mean, new variance is 100X the old variance (square result to get st dev). And it is similarly straightforward to determine the probability of being X st devs above the mean (works just like the above example).

But here is my question, how do I determine the probability of the result being > x st devs above the mean at any point in the 100 trial run, not just at the end?

I mean I could calculate the probability for 1 run, 2 runs, 3 runs, etc up to 100, but I see no way to derive the cumulative probability that it happens at any point during that run.

Any ideas?

Let me know if you need any clarification of my question.

[ncray]

First of all, if X~N(10, 25), then the probability that X > 15 is given by integrating the pdf from 15 to infinity, giving you ~ 0.158655. Since we know that for a normal distribution, about 68% of the mass will lie within one standard deviation within the mean, (1-.68)/2 gives a decent approximation.

What is your random variable for 100 trials? If you want the new mean to be 100 * the old mean, then let me define a RV Y = X_1 + ... + X_100, where X_1,...X_100 are iid ~ N(10, 25). Then E[Y] = E[X_1] + ... + E[X_100] = 100 * 10 = 1000.

Var[Y] = Var[X_1 + ... + X_100] by independence = Var[X_1] + ... + Var[X_100] = 100*25 = 2500.

You need to know that the sum of iid normal random variables is also normal, so Y~N[100, 2500].

If you want to stop at any point in the 100 trial run, let Z_n = sum from i = 1 to n of the X_i's. Then Z~[n*mu, n*sigma^2] where mu is the mean of the X_i's and sigma^2 is the variance of the X_i's.

In order to get the probability of a result being x st devs above the mean, integrate the resulting pdf from mean + x*(the square root of n*sigma^2) to infinity.

[alThor]

You are asking a Risk of Ruin question. In some cases, the probability you are looking for is roughly twice the likelihood that you end beyond X standard deviations from the mean. I'll let the ROR specialists expand on this.