Random number generation: probability of 2nd smallest >0.002

[psie]

Homework Statement

One hundred numbers, uniformly distributed in the interval $(0,1)$ are generated by a computer. What is the probability that the largest number is at most $0.9$? What is the probability that the second smallest number is at least $0.002$?

Relevant Equations

The extreme order variables $X_{(1)}=\min\{X_1,\ldots,X_{100}\}$ and $X_{(100)}=\max\{X_1,\ldots,X_{100}\}$ have cdf $F_{X_{(1)}}(x)=1-(1-F(x))^{100}$ and $F_{X_{(100)}}(x)=(F(x))^{100}$ respectively, where $F$ is the cdf of the iid $X_1,\ldots,X_{100}$.

The first question is fairly straightforward. The density of $X$ (i.e. one of the iid r.v.s. $X_1,\ldots,X_{100}$) is just $f(x)=1$ for $0<x<1$ and $0$ otherwise. The cdf $F$ is therefore $F(x)=x$ for $0<x<1$, $F(x)=0$ for $x<0$ and $F(x)=1$ for $x>1$. In the first question, we are interested in \begin{align*} P(X_{(100)}<0.9)&=F_{X_{(100)}}(0.9) \\ &=(0.9)^{100}.\end{align*}For the 2nd question, I don't know how to approach this and I'm stuck. The cdf of arbitrary order variables has not been derived yet. I know we are looking for $P(X_{(2)}>0.002)=1-P(X_{(2)}<0.002)$. But maybe there's a workaround. I only know the formulas in the relevant equations above. Grateful for any help.

 

[BvU]

Hi,

so the condition in the second question is met in two cases:
a) if all 100 numbers are > 0.002 -- which has a probability ...
b) if one number is < 0.002 AND all others are > 0.002

one number is < 0.002 has probability ...​

99 others > 0.002 probability ...​

$\$

 

[psie]

Hi,

so the condition in the second question is met in two cases:
a) if all 100 numbers are > 0.002 -- which has a probability ...
b) if one number is < 0.002 AND all others are > 0.002

one number is < 0.002 has probability ...​

99 others > 0.002 probability ...​

$\$

Ok. So (a) occurs with probability \begin{align*}P(X_1>0.002,X_2>0.002,\ldots,X_n>0.002)&=\prod_{i=1}^{100} (1-P(X_i<0.002))\\ &=(1-F(0.002))^{100} \\ &=(0.998)^{100}\approx 0.819.\end{align*} But for (b), I am not sure how to proceed. We've got $100$ possibilities for one of the sample points to be less than $<0.002$ and all others $>0.002$, right? Do we just multiply $100$ with $$P(X_1>0.002,X_2>0.002,\ldots,X_k<0.002,\ldots, X_n>0.002)?$$ If one has obtained the probability for (b), I am not sure how one obtains the (total) probability for the second question.

 

[psie]

Ok, I think I know now how to calculate the probability in (b). We are essentially making 100 trials and looking for success (i.e. that a number is greater 0.002) in 99 of the trials. The success probability is $$P(X>0.002)=1-P(X<0.002)=1-F(0.002)=0.998.$$ Now, the probability that the count $Y=99$ follows a binomial distribution, so $$P(Y=99)=\binom{100}{99} (0.998)^{99}(0.002)^{100-99}\approx 0.164.$$As (a) and (b) are disjoint events, we have that the probability for the 2nd question is $$0.164+0.819=0.983.$$That seems like a very high probability.

 

[BvU]

I agree it's rather high.
I painted myself in a corner trying to get the result coming from the other side: there are $\binom{100}{2}=4950$ possible pairs. Require both $\ <0.002$ gives $0.0198$ --- which is not $1-0.9826 = 0.0174$.
Oh boy ....

$\$

 

[pbuk]

trying to get the result coming from the other side: there are $\binom{100}{2}=4950$ possible pairs. Require both $\ <0.002$ gives $0.0198$

So that is the probability that exactly 2 numbers are less than 0.002. What happens if exactly 3 numbers are less than 0.002?

 

[psie]

I think the answer I got is correct.

We can check it via Wikipedia and WolframAlpha. The density of the 2nd order variable $U_{(2)}$ (see Wikipedia) is $$f_{U_{(2)}}(x) = \frac{100!}{98!} x (1-x)^{98}.$$Integrating from $0.002$ to $1$ gives (see WolframAlpha) approximately $0.983$.

 

[BvU]

So that is the probability that exactly 2 numbers are less than 0.002. What happens if exactly 3 numbers are less than 0.002?

Yep, the check I tried was too hasty : forgot the (1-0.002)^98 and overlooked the 3, 4, 5, etc.
Good catch !

$\$

 

[pbuk]

I think the answer I got is correct.

So do I, although perhaps a little over-complicated getting there: you seem to have focussed on PDFs and CDFs but both parts of the problem can be more simply answered from binomial fundamentals.

What is the probability that the largest number is at most $0.9$?

This is the probability that all 100 numbers are less than 0.9: $0.9^{100}$.

What is the probability that the second smallest number is at least $0.002$?

This is the probability that either:
- all 100 numbers are greater than 0.002: $0.998^{100} \approx 0.8186$; or
- exactly 1 number is less than 0.002: $\binom{100}{1} 0.002^1 0.998^{99} \approx 0.1640$.
Adding the two (they are as you say mutually exclusive) gives the answer
$$ 0.998^{100} + \binom{100}{1} 0.002^1 0.998^{99} \approx 0.9826 $$

Note that you can often boost your confidence in an answer with a quick numerical simulation.