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Given samples from two empirical distributions (not necessarily passing tests of normality, but a solution for the normal case would definitely be useful) what's the probability that the maximum value from sample 1 will be greater than the maximum value from sample 2?

I have methods for approximating the probability by brute force computation using Mathematica's built in empirical distribution / probability handling, but I would like a less opaque way to cross check my results.

To give some context - imagine you are proposed with the following:

  • The average height of citizens in Country A is 72 inches with a standard deviation of 6 inches
  • The average height of citizens in country B is 66 inches with a standard deviation of 3 inches
  • A random sample of 5 citizens is drawn from country A
  • A random sample of 25 citizens is drawn from country B

Question: What is the probability that the tallest member of sample A is taller than the tallest member of sample B?

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  • $\begingroup$ Hint: consider the problem in the other direction. Suppose you drew a random sample of 25 citizens from country B. What would the sample standard deviation have to be for you to estimate the total deviation as 3 inches? How does the standard deviation of a sample compare to the standard deviation of the population, given a number of samples n? $\endgroup$ – barrycarter Mar 16 '16 at 23:43
  • $\begingroup$ Are the elements sampled independently? $\endgroup$ – Vossler Mar 17 '16 at 2:53
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    $\begingroup$ sjsu.edu/faculty/watkins/unboundedmax.htm may be helpful. @Vossler uses those formulas below, but I had to think about them for a second: the chance that the max of n samples is x is the chance that n-1 samples are < x and 1 sample equals x. To find this, we integrate CDF^(n-1)(x) * PDF(x), as Vossler does below. $\endgroup$ – barrycarter Mar 17 '16 at 5:04
  • $\begingroup$ I'm working on this and got an approximate answer of about .9515 for your specific case. Is this close to what you're getting? $\endgroup$ – barrycarter Mar 17 '16 at 17:44
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    $\begingroup$ Are you familiar with Mathematica's "ExtremeValueDistribution"? $\endgroup$ – barrycarter Mar 18 '16 at 16:11
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I hope I got your question right and your question is mathematical in the sense that you want to get an answer in terms of the distribution of the sample.

Assume you have two samples $X_1,...,X_n\sim X$ and $Y_1,...Y_n\sim{Y}$. If the two samples are from the same distribution you can just take $X=Y$, this doesn't matter here.

Since you seem to be motivated by a practical application, I will assume that that both $X$ and $Y$ are absolutely continuous (i. e. have densities $p_x, p_y$). Let $M_x^n=\max_{i=1...n}X_i, M_y^m=\max_{i=1...m}Y_i$.

First, for any iid sample from $Y$ we have $$ P(M_y^m<t)=\prod_{i=1}^m P(Y_i<t)=(F_y(t))^m $$ and the corresponding density is $$ p_{M_y^m}(t)=mF_y^{m-1}(t)p_y(t). $$

Now,

$$ P(M_x^n<M_y^m)=\int P(M_x^n<t)p_{M_y^m}(t)dt=\int F_{M_x^n}(t)p_{M_y^m}(t)dt=\int m(F_x(t))^n (F_y(t))^{m-1}p_y(t)dt. $$

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NOTE: This answer assumes the heights are normally distributed.

I thought there'd be a well-known distribution for the maximum of n standard normal variables, but, as Expected value for maximum of n normal random variable notes, there isn't. You can't even find a closed form for the mean/expected value (though you can for the median).

In Mathematica, this distribution is: "OrderDistribution[{NormalDistribution[0, 1], n}, n]"

The PDF for the max of n standard normal variables is:

$ \frac{2^{\frac{1}{2}-n} n e^{-\frac{x^2}{2}} \left(\text{erf}\left(\frac{x}{\sqrt{2}}\right)+1\right)^{n-1}}{\sqrt{\pi }} $

and the CDF is:

$2^{-n} \left(\text{erf}\left(\frac{x}{\sqrt{2}}\right)+1\right)^n$

Here's what the PDF looks like for various values of n:

enter image description here

Of course, n=1 is just the standard normal distribution itself.

Except for n=1, these aren't normal distributions, but can be approximated as such. For n=5, for example,
$\{\mu \to 1.11241,\sigma \to 0.656668\}$ and the graph looks like this:

enter image description here

The integral of the difference squared from $-\infty$ to $\infty$ is about 0.000848615

For n=25, $\{\mu \to 1.89997,\sigma \to 0.486849\}$:

enter image description here

with an error squared of 0.00331485.

In both cases, I used numerical methods to minimize integral of the difference squared (ie, the "error" squared).

Of course, you used 5 and 25 as examples. Below is a tabulation for the best fit parameters from 1 to 25, plus some larger values for comparison (n=1 is the standard normal distribution):

$ \begin{array}{ccccc} \text{n} & \mu _{\text{fit}} & \sigma _{\text{fit}} & \text{err}^2 & \text{} \\ 1 & 0 & 1 & 0 & \text{} \\ 2 & \text{ 0.536} & \text{ 0.820} & \text{ 0.000145} & \text{} \\ 3 & \text{ 0.806} & \text{ 0.740} & \text{ 0.000382} & \text{} \\ 4 & \text{ 0.983} & \text{ 0.691} & \text{ 0.000622} & \text{} \\ 5 & \text{ 1.112} & \text{ 0.657} & \text{ 0.000849} & \text{} \\ 6 & \text{ 1.214} & \text{ 0.631} & \text{ 0.001058} & \text{} \\ 7 & \text{ 1.297} & \text{ 0.611} & \text{ 0.001252} & \text{} \\ 8 & \text{ 1.366} & \text{ 0.595} & \text{ 0.001432} & \text{} \\ 9 & \text{ 1.427} & \text{ 0.582} & \text{ 0.001599} & \text{} \\ 10 & \text{ 1.479} & \text{ 0.570} & \text{ 0.001755} & \text{} \\ 11 & \text{ 1.526} & \text{ 0.560} & \text{ 0.001901} & \text{} \\ 12 & \text{ 1.568} & \text{ 0.551} & \text{ 0.002038} & \text{} \\ 13 & \text{ 1.606} & \text{ 0.543} & \text{ 0.002167} & \text{} \\ 14 & \text{ 1.641} & \text{ 0.536} & \text{ 0.002289} & \text{} \\ 15 & \text{ 1.673} & \text{ 0.529} & \text{ 0.002405} & \text{} \\ 16 & \text{ 1.703} & \text{ 0.524} & \text{ 0.002515} & \text{} \\ 17 & \text{ 1.730} & \text{ 0.518} & \text{ 0.002619} & \text{} \\ 18 & \text{ 1.756} & \text{ 0.513} & \text{ 0.002719} & \text{} \\ 19 & \text{ 1.780} & \text{ 0.509} & \text{ 0.002815} & \text{} \\ 20 & \text{ 1.803} & \text{ 0.504} & \text{ 0.002907} & \text{} \\ 21 & \text{ 1.825} & \text{ 0.500} & \text{ 0.002995} & \text{} \\ 22 & \text{ 1.845} & \text{ 0.497} & \text{ 0.003079} & \text{} \\ 23 & \text{ 1.864} & \text{ 0.493} & \text{ 0.003161} & \text{} \\ 24 & \text{ 1.882} & \text{ 0.490} & \text{ 0.003239} & \text{} \\ 25 & \text{ 1.900} & \text{ 0.487} & \text{ 0.003315} & \text{} \\ 50 & \text{ 2.182} & \text{ 0.440} & \text{ 0.004654} & \text{} \\ 100 & \text{ 2.440} & \text{ 0.403} & \text{ 0.006047} & \text{} \\ 500 & \text{ 2.971} & \text{ 0.342} & \text{ 0.009231} & \text{} \\ 1000 & \text{ 3.176} & \text{ 0.323} & \text{ 0.010527} & \text{} \\ 5000 & \text{ 3.616} & \text{ 0.287} & \text{ 0.013320} & \text{} \\ 10000 & \text{ 3.791} & \text{ 0.275} & \text{ 0.014434} & \text{} \\ 100000 & \text{ 4.328} & \text{ 0.244} & \text{ 0.017810} & \text{} \\ \end{array} $

I couldn't get Mathematica to compute best fit mean and sd for values much higher than n=100000, but I didn't try that hard.

I couldn't find a closed form for the best fit mean, or even the true mean, but there is one for the median:

$\sqrt{2} \text{erf}^{-1}\left(2^{\frac{n-1}{n}}-1\right)$

Since we know the PDF, you'd think we could find a closed form for the mode, by setting the PDF's derivative to 0, which would be:

$ \frac{2^{-n} e^{-x^2} \left(\text{erf}\left(\frac{x}{\sqrt{2}}\right)+1\right)^{n-2} \left(\sqrt{2 \pi } n e^{\frac{x^2}{2}} x \left(\text{erfc}\left(\frac{x}{\sqrt{2}}\right)-2\right)+2 (n-1) n\right)}{\pi }=0 $

As it turns out (https://mathematica.stackexchange.com/questions/110564/) this can be solved for n, but not for x. Thus, I use numerical techniques to obtain that as well.

Below are the various "means" (including the best-fit mean above) for various values.

$ \begin{array}{ccccc} \text{n} & \mu _{\text{fit}} & \mu _{\text{true}} & \text{Median} & \text{Mode} \\ 1 & 0 & 0 & 0 & 0 \\ 2 & \text{ 0.536} & \text{ 0.564} & \text{ 0.545} & \text{ 0.506} \\ 3 & \text{ 0.806} & \text{ 0.846} & \text{ 0.819} & \text{ 0.765} \\ 4 & \text{ 0.983} & \text{ 1.029} & \text{ 0.998} & \text{ 0.936} \\ 5 & \text{ 1.112} & \text{ 1.163} & \text{ 1.129} & \text{ 1.062} \\ 6 & \text{ 1.214} & \text{ 1.267} & \text{ 1.231} & \text{ 1.160} \\ 7 & \text{ 1.297} & \text{ 1.352} & \text{ 1.315} & \text{ 1.241} \\ 8 & \text{ 1.366} & \text{ 1.424} & \text{ 1.385} & \text{ 1.309} \\ 9 & \text{ 1.427} & \text{ 1.485} & \text{ 1.446} & \text{ 1.368} \\ 10 & \text{ 1.479} & \text{ 1.539} & \text{ 1.499} & \text{ 1.420} \\ 11 & \text{ 1.526} & \text{ 1.586} & \text{ 1.546} & \text{ 1.466} \\ 12 & \text{ 1.568} & \text{ 1.629} & \text{ 1.588} & \text{ 1.508} \\ 13 & \text{ 1.606} & \text{ 1.668} & \text{ 1.626} & \text{ 1.545} \\ 14 & \text{ 1.641} & \text{ 1.703} & \text{ 1.662} & \text{ 1.580} \\ 15 & \text{ 1.673} & \text{ 1.736} & \text{ 1.694} & \text{ 1.611} \\ 16 & \text{ 1.703} & \text{ 1.766} & \text{ 1.724} & \text{ 1.641} \\ 17 & \text{ 1.730} & \text{ 1.794} & \text{ 1.751} & \text{ 1.668} \\ 18 & \text{ 1.756} & \text{ 1.820} & \text{ 1.777} & \text{ 1.693} \\ 19 & \text{ 1.780} & \text{ 1.844} & \text{ 1.801} & \text{ 1.717} \\ 20 & \text{ 1.803} & \text{ 1.867} & \text{ 1.824} & \text{ 1.740} \\ 21 & \text{ 1.825} & \text{ 1.889} & \text{ 1.846} & \text{ 1.761} \\ 22 & \text{ 1.845} & \text{ 1.910} & \text{ 1.866} & \text{ 1.781} \\ 23 & \text{ 1.864} & \text{ 1.929} & \text{ 1.885} & \text{ 1.800} \\ 24 & \text{ 1.882} & \text{ 1.948} & \text{ 1.904} & \text{ 1.819} \\ 25 & \text{ 1.900} & \text{ 1.965} & \text{ 1.921} & \text{ 1.836} \\ 50 & \text{ 2.182} & \text{ 2.249} & \text{ 2.204} & \text{ 2.117} \\ 100 & \text{ 2.440} & \text{ 2.508} & \text{ 2.462} & \text{ 2.375} \\ 500 & \text{ 2.971} & \text{ 3.037} & \text{ 2.992} & \text{ 2.908} \\ 1000 & \text{ 3.176} & \text{ 3.241} & \text{ 3.198} & \text{ 3.115} \\ 5000 & \text{ 3.616} & \text{ 3.678} & \text{ 3.636} & \text{ 3.558} \\ 10000 & \text{ 3.791} & \text{ 3.852} & \text{ 3.811} & \text{ 3.735} \\ 100000 & \text{ 4.328} & \text{ 4.384} & \text{ 4.346} & \text{ 4.276} \\ \end{array} $

Of course, if these were true normal distributions, the median, mode, mean, and "best fit $\mu$" would be identical.

You can estimate the standard deviation by comparing the distribution to the normal distribution in the limiting case around the median. This is only estimate I could find with a closed form:

$ \frac{2^{\frac{n-1}{n}} e^{\text{erf}^{-1}\left(2^{\frac{n-1}{n}}-1\right)^2}}{n} $

Below is a table of the best fit standard deviation, the true standard deviation (computed numerically), and the closed-form estimated standard deviation above:

$ \begin{array}{cccc} \text{n} & \sigma _{\text{fit}} & \sigma _{\text{true}} & \sigma _{\text{est}} \\ 1 & 1 & 1 & 1 \\ 2 & \text{ 0.820} & \text{ 0.826} & \text{ 0.820} \\ 3 & \text{ 0.740} & \text{ 0.748} & \text{ 0.740} \\ 4 & \text{ 0.691} & \text{ 0.701} & \text{ 0.692} \\ 5 & \text{ 0.657} & \text{ 0.669} & \text{ 0.659} \\ 6 & \text{ 0.631} & \text{ 0.645} & \text{ 0.634} \\ 7 & \text{ 0.611} & \text{ 0.626} & \text{ 0.614} \\ 8 & \text{ 0.595} & \text{ 0.611} & \text{ 0.598} \\ 9 & \text{ 0.582} & \text{ 0.598} & \text{ 0.585} \\ 10 & \text{ 0.570} & \text{ 0.587} & \text{ 0.574} \\ 11 & \text{ 0.560} & \text{ 0.577} & \text{ 0.564} \\ 12 & \text{ 0.551} & \text{ 0.569} & \text{ 0.555} \\ 13 & \text{ 0.543} & \text{ 0.561} & \text{ 0.548} \\ 14 & \text{ 0.536} & \text{ 0.555} & \text{ 0.541} \\ 15 & \text{ 0.529} & \text{ 0.549} & \text{ 0.534} \\ 16 & \text{ 0.524} & \text{ 0.543} & \text{ 0.529} \\ 17 & \text{ 0.518} & \text{ 0.538} & \text{ 0.523} \\ 18 & \text{ 0.513} & \text{ 0.533} & \text{ 0.519} \\ 19 & \text{ 0.509} & \text{ 0.529} & \text{ 0.514} \\ 20 & \text{ 0.504} & \text{ 0.525} & \text{ 0.510} \\ 21 & \text{ 0.500} & \text{ 0.521} & \text{ 0.506} \\ 22 & \text{ 0.497} & \text{ 0.518} & \text{ 0.502} \\ 23 & \text{ 0.493} & \text{ 0.514} & \text{ 0.499} \\ 24 & \text{ 0.490} & \text{ 0.511} & \text{ 0.496} \\ 25 & \text{ 0.487} & \text{ 0.508} & \text{ 0.493} \\ 50 & \text{ 0.440} & \text{ 0.464} & \text{ 0.447} \\ 100 & \text{ 0.403} & \text{ 0.429} & \text{ 0.411} \\ 500 & \text{ 0.342} & \text{ 0.370} & \text{ 0.351} \\ 1000 & \text{ 0.323} & \text{ 0.351} & \text{ 0.332} \\ 5000 & \text{ 0.287} & \text{ 0.316} & \text{ 0.297} \\ 10000 & \text{ 0.275} & \text{ 0.304} & \text{ 0.285} \\ 100000 & \text{ 0.244} & \text{ 0.272} & \text{ 0.253} \\ \end{array} $

As you can see, you can kinda-sorta estimate the best fit mean and standard deviation based on the closed forms we found earlier.

If we use the best fit approximations, your distributions are:

  • tallest of 5 from Country A: $\{\mu \to 78.6745,\text{sd}\to 3.94001\}$

  • tallest of 25 from Country B: $\{\mu \to 71.6999,\text{sd}\to 1.46055\}$

  • The difference: $\{\mu \to 6.9746,\sigma \to 4.20201\}$ (based on addition of variances)

So the chance A will be taller is about 95.15%

Of course, that's just an approximation, so let's do a more rigorous analysis below.

The CDF for the tallest of 25 from Country B:

$ \frac{\left(\text{erf}\left(\frac{x-66}{3 \sqrt{2}}\right)+1\right)^{25}}{33554432} $

and the PDF for the tallest of 5 from Country A:

$ \frac{5 e^{-\frac{1}{72} (x-72)^2} \left(\text{erf}\left(\frac{x-72}{6 \sqrt{2}}\right)+1\right)^4}{96 \sqrt{2 \pi }} $

If we use Mathematica to numerically integrate the product from $-\infty$ to $\infty$ we get 95.77%, which is pretty close to our earlier estimate.

Just for fun, I ran a Monte Carlo simulation as well:

 
countryA := Max[RandomVariate[NormalDistribution[72, 6], 5]] 
countryB := Max[RandomVariate[NormalDistribution[66, 3], 25]] 
t =Table[countryA>countryB,{i,1,100000}] 

and got 95.79% (I ran it several times, 95.79% is the average).

Mathematica derivations/formulas for this answer can be found at: https://github.com/barrycarter/bcapps/blob/master/STACK/bc-prob-max.m

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