0

I'm trying to solve the following question:

Let $Y_1, Y_2, \ldots , Y_n$ be independent and uniformly distributed R.V on $[0, 1]$. Let $x, x \in [0, 1]$, be fixed and write $N = \min\{k \ge 1 | Y_1 + Y_2 + \dots + Y_k > x\}$. What is the probability of the event $N = n$.

My solution is as the following:

The probability of the event $N = n$:

$P(Y_1+Y_2+\dots+Y_n > x)$ AND $P(Y_1+Y_2+\dots +Y_{n-1} < x)$ AND $P(Y_1+Y_2+\dots +Y_{n-2} + Y_n < x) \dots $

Where in the second term, I excluded $Y_n$ and in the third term I excluded $Y_{n-1}$. I am not sure whether I should do this permutation or I should just stick with the following solution:

The probability of the event $ N = n$:

$P(Y_1+Y_2+ \dots +Y_n > x)$ AND $P(Y_1+Y_2+\dots +Y_{n-1} < x)$

If $Y$'s are not random variables and they represent drawing a ball that has a certain number, I think the first solution is the way to go, isn't it? However, in the case of the random variable, I'm not sure what the correct way.

kjetil b halvorsen
  • 63,378
  • 26
  • 142
  • 467
cyberic
  • 123
  • 1
  • 9
  • 1
    See https://stats.stackexchange.com/questions/194352/. You don't seem to have characterized the event $N=n$ as intended: it is the event $$Y_1 + \cdots + Y_{n-1} \le x\lt Y_1+\cdots+Y_{n-1}+Y_n .$$ – whuber Feb 03 '18 at 17:20
  • @whuber In this case, I think the question should explicitly say that the Y's are "drawing" sequentially. In other words, Yn-2 cannot be followed by Yn. Right? – cyberic Feb 03 '18 at 17:27
  • The question is explicit in that it has already indexed the $Y$'s once and for all. There's no reason why it should repeat what has already been made plain by the notation. – whuber Feb 03 '18 at 17:30
  • @whuber What do you mean by "once and for all"? and which notation are you referring to? I believe I'm missing something fundamental in such type of questions – cyberic Feb 03 '18 at 17:35
  • I refer to *your* notation. Suppose $n=3$. Instead of calling the results $Y_1, Y_2, Y_3$ you could just as well call them Abe, Bob, and Cyd, respectively. In these terms the question states, "Let $N=1$ if Abe exceeds $x$, let $N=2$ if otherwise Abe+Bob exceeds $x$, and let $N=3$ if otherwise Abe+Bob+Cyd exceeds $x$. What is the chance $N=3$?" Incidentally, notice that $N$ is not completely defined: we aren't told what it equals when Abe+Bob+Cyd fails to exceed $x$: maybe it equals $3$ in that case, or maybe not. Strictly speaking, the answer is that the chance of this event is indeterminate. – whuber Feb 03 '18 at 18:59

0 Answers0