Suppose I have the following problem, I want to add a random uniform number from $[0,1]$ every time in a variable. If n is the number of times I added, then:
$$ X(n) = \sum_{i=0}^{n} U_i $$
Where $U_i$ are uniform random variables from $[0,1]$.
I want to for each n the probability of $X$ exceeding $0.3$ at exactly that n.
In other words, I want to find the following probability:
$$ P(X(n)>0.3 | X(n-1)<0.3) $$
I now $X(n)$ is an Irwin-Hall distribution but I 'm struggling because $X(n)$ depends on $X(n-1)$ and so it is incorrect to say that:
$$ P(X(n)>0.3 | X(n-1)<0.3) = P(X(n)>0.3)*P(X(n-1)<0.3) $$
The probability of $X(1)$ first being greater than $z$ (with $0<z<1$) will be $1-z$. That, of course, you already know.
So we need to find the probability that $X(n)$ is the first time that sum is greater than $z$. The probability that $X(n-1)<z$ and $X(n)>z$ requires integrating over the product of the pdf of the Irwin-Hall distribution for $n-1$ and the probability that the $n$-th observation kicks the sum over $z$.
For $n>0$ and $0<z<1$, the pdf of the Irwin-Hall distribution is
$${z^{n-1}}\over{(n-1)!}$$
So the integral of interest is
$$\int_0^z {{x^{n-2}}\over{(n-2)!}}(1-(z-x))=\frac{(n-z) z^{n-1}}{n!}$$
For $n=1$ and $z=0.3$ this results in 0.7. For $n=2$ and $z=0.3$ this results in 0.255.