Probability of Sum of Independent Events Exceeding a Value

Question

Suppose I have $n$ random number generators. Once an hour, on the hour, each one generates a random real number $x_k$ such that $0 \le x_k \lt \infty$. Each generator produces its values according to its own independent probability distribution function $f_k()$, which is a known function. For example, one generator might follow an exponential distribution, another might follow a normal distribution, etc.

Let $X = \sum\limits_{k=1}^n x_k$ for all of the number generators in any one hour.

Given $y$ such that $0 \le y \lt 1$ (a probability), I need to find a value $z$ such that $P(X \le z) = y$.

Basically, I need to be able to do something like find the value that $X$ will be less than or equal to 50% of the time.

I apologize if I've gotten any of the notation wrong, I'm actually a software engineer so I know some things about math but not others. I know enough about probability to express the problem above, but I don't even know where to begin in terms of solving it. Any help, or even suggested readings would be much appreciated.

Answer 1

Suppose they are all generating uniformly distributed random numbers in the closed interval $[a,b]$. Then your sum will never exceed $nb$ whereas that wouldn't be the case if they were generating normally distributed random numbers. The way you pose the question is way too generic. You need to be more specific about the distribution of each constituent random variable in your sum.

If $n$ is large, you can invoke the Central Limit Theorem. Your random variables don't have to be identically distributed (independence is sufficient). That way you only need to know the mean and variance of random variable in the sum to calculate probabilities since the sum will be approximately normal.

Answer 2

Well for start, if the distribution are poissons, then also the sum of the variables will have poisson distribution, with the parameter equal to the sum of the parameters.

Otherwise (if $X_i$ admit densities $f_i$) since they are independent then the random vector $Z = (X_1, \dots, X_n)$ admits a density $f_Z (x_1, \dots , x_n) = f_1(x_1) \cdot \cdot \cdot f_n(x_n)$

Then one can find the distribution of $X_1 + \dots + X_n$ applying Jacobi transformation formula with the function $g(x) = (x_1,x_2, \dots, x_{n-1}, x_1 + \dots + x_n)$

This way you can find the distribution of $(X_1,X_2, \dots, X_{n-1}, X_1 + \dots + X_n)$ and you can calculate the marginal density (of the last element of the vector ) in the usual way.

That said, it is unlikely that you will be able to compute all those steps in a general case, because the calculations tend to become very difficult.