/sci/ - Science & Math

>>12341872
I'm assuming you mean "I draw points uniformly distributed in [math][a, b][/math] and stop drawing when the sum surpasses some [math]x[/math], how many points do I draw on average?"
If [math]a > 0[/math], you surpass the value in at most [math]x/a[/math] draws. Then you can just compute the probability of stopping at each [math]n[/math] on the way and compute the expectation normally. I recommend using a computer.
If [math]a \leq 0[/math], the situation changes. I'll scale things so that [math]b = 1[/math], and I'll assume [math]a=0[/math] for reasons to be explained later.
Then, the probability of drawing [math]n[/math] balls and staying below [math]x[/math] is [math]min \left ( \frac{x}{n}, 1 \right )[/math]. I'll assume [math]n[/math] is very big, so the formula is just [math]\frac{x}{n}[/math].
Then, notice that you stop at [math]n[/math] if and only if you fail at [math]n-1[/math], which happens with probability [math]\frac{x}{n-1}[/math], and do not fail at [math]n[/math], which has a probability of [math]\frac{x}{n}[/math]. Hence you have a probability of [math]\frac{x}{n-1} - \frac{x}{n} = \frac{x}{n(n-1)}[/math] of stopping at [math]n[/math].
When we try to compute the expectation from this, we get a term [math]\sum _{n = N}^{\infty} \frac{nx}{n(n-1)} = \sum _{n=N}^{\infty}\frac{x}{n-1}[/math], but that's a harmonic series and diverges. Thinking for two seconds tells you that this also happens when [math]a < 0[/math].