Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

choose a random number between 0 and 1 and record its value. Do this again and add the second number to the first number. Keep doing this until the sum of the numbers exceeds 1. What's the expected valued of the number of random numbers needed to accomplish this?

share|improve this question
    
What have you tried? Is the random number chosen uniformly in $[0,1]$? –  Najib Idrissi Feb 20 '12 at 16:36
3  
See mathworld.wolfram.com/UniformSumDistribution.html Scroll down until you hit the reference to Derbyshire –  Byron Schmuland Feb 20 '12 at 16:49

3 Answers 3

Here is a (perhaps) more elementary method. Let $X$ be the amount of numbers you need to add until the sum exceeds $1$. Then (by linearity of expectation) $$ \mathbb{E}[X] = 1 + \sum_{k \geq 1} \Pr[X > k]. $$ Now $X > k$ if the sum of the first $k$ numbers $x_1,\ldots,x_k$ is smaller than $1$. This is exactly equal to the volume of the $k$-dimensional set $$ \{(x_1,\ldots,x_k) : \sum_{i=1}^k x_i \leq 1, \, x_1,\ldots,x_k \geq 0\}. $$ This is known as the $k$-dimensional simplex. When $k = 1$, we get a line segment of length $1$. When $k = 2$, we get a right equilateral triangle with sides of length~$1$, so the area is $1/2$. When $k=3$, we get a triangular pyramid (tetrahedron) with unit sides, so the volume is $1/6$. In general, the volume is $1/k!$, and so $$ \mathbb{E}[X] = 1 + \sum_{k \geq 1} \frac{1}{k!} = e. $$

share|improve this answer
    
we haven't learned expected value yet. There must be another way... –  user25329 Feb 20 '12 at 18:17
5  
@user25329: You're asked to compute the expected value, but you haven't learned what that is yet? What? –  BlueRaja - Danny Pflughoeft Feb 20 '12 at 18:35
    
yeah it sounds absurd but we haven't...but this is a "bonus" problem... –  user25329 Feb 20 '12 at 19:18

Assuming the numbers come from a uniform distribution over $[0,1]$ and that the trials are independent, here is an outline (this is example 7.4 4h. in Sheldon Ross' A First Course in Probability, sixth edition):

1) Let $X_i$ be the number obtained on the $i$'th trial.

2) For $0\le x\le1$, let $Y(x)$ be the minimum number of trials needed so that the sum of the $X_i$ exceeds $x$. Set $e(x)=\Bbb E [Y(x)]$.

3) Compute $e(x)$ by conditioning on the value of $X_1$: $$\tag{1} e(x)=\int_0^1 \Bbb E [ Y(x) | X_1=y]\, dy. $$ Here, use the fact that $$\tag{2}\Bbb E [ Y(x) | X_1=y] = \cases{1,& $y>x$\cr 1+e(x-y),& $y\le x $}.$$

Substitution of $(2)$ into $(1)$ will give $$\tag{3} e(x)=1+\int_0^x e(u)\,du. $$

4) Solve equation $(3)$ (by differentiating both sides with respect to $x$ first) for $e(x)$.

5) You wish to find $e(1)$.

share|improve this answer
    
Can you please put it into simpler terms?..I am only in PreCal I haven't learned a lot of the symbols you used, do you mind to explain? Thanks! –  user25329 Feb 20 '12 at 17:07
1  
@user25329 I'm sorry, but I don't see how to solve this problem without using results from calculus. Perhaps I misinterpreted your question. When you say "between 0 and 1", do you mean any number in the interval $[0,1]$ or did you mean either 0 or 1? The symbols I used were $\Bbb E$ for expected value, so $\Bbb E[(Y(x)]$ is the expected number of "number choosing" so that the sum exceeds $x$. $\int$ is the integration sign from calculus. Equation $(1)$ is a formula used for calculating conditional expectations. If you haven't seen these concepts before, you should ignore my answer :) –  David Mitra Feb 20 '12 at 17:18
    
No you didn't misinterpret my question. But I haven't seen these terms before I think we are expected to solve it in another way I guess. But thank you so much for providing such detailed answers to my question. Thank you! –  user25329 Feb 20 '12 at 17:28

Here is another method. This is exactly the same technique Yuval has given above but with an added step that goes through computing the pmf first before computing the expectation. Hopefully, this will help you understand the problem from a slightly different angle.

Let us denote by $N$ the number of random variables we need to add for the sum to exceed $1$. We first find the distribution (probability mass function) of $N$. The easiest way to do this is by computing $P(N > n)$ for $n=1,2,3,\dots$. Once we know this, we can compute $P(N = n) = P(N > n-1) - P(N > n)$.

The event that $(N > n)$ in plain English says that the sum of the first $n$ uniform random variables did not exceed $1$. The probability of this event is therefore $P(N > n) = P(U_1 + U_2 + \dots + U_n < 1)$. This can be calculated by a standard multi-dimensional integral to be $\frac{1}{n!}$. If you don't want to use calculus, one can justify this result geometrically but integration is perhaps the most natural approach to evaluate this probability. Therefore, we have $P(N = n) = \frac{1}{(n-1)!} - \frac{1}{n!} = \frac{n-1}{n!}$ for $n=1,2,3\dots$.

Once we know the pmf of N, we calculate

$E(N) = \sum_{n=1}^{\infty} n P(N=n) = \sum_{n=1}^{\infty} \frac{n(n-1)}{n!} = \sum_{n=0}^{\infty} \frac{1}{n!} = e$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.