#standard normal

So the aliens have returned, and the goal here is to convert them to our weird religion of “geometry,” to explain to them why that “pi” or “tau” thing is actually important.

One involves probability.  The aliens understand Bernoulli trials, which we’ll simplify to be a fair coin, 50/50.  Let’s go back to that definition of pi as an integral - namely, twice the integral of (1+x^2)^(0.5).  If you could just replace 0.5 with an integer, it would be easy to work out, even before calculus as we know it was known - Cavalieri’s quadrature formula lets you do that.  You can just multiply it out and play with all the integral exponents.  Ten years before the annus mirabilis, John Wallis lamented this fact (he was not, as claimed here, “thinking about sine waves,” the real significance of which hadn’t been established in his day) and, by considering the relationship of each such integral to the next, came up with a workaround.

So what does this have to do with fair coins?  Well, let’s bear in mind that the string (which I just got from random.org) THTHTTHTHT is exactly as likely as... um... that doesn’t look very random.  Another one.  THHTHHTTTT.  Good enough.  The point is that each of those are exactly as likely as ten heads or ten tails.  The reason they feel more likely... if they do feel more likely... is that there are 210 ways to get four heads and six tails, and only one way each to get ten heads or ten tails.

Before generalizing this, let’s consider the fact that mathematicians can instantly identify fake randomness by the fact that it’s too close to alternating, or too close to being 50/50.  I don’t know if de Moivre knew this, but he posed the question, what are the chances of getting exactly 50/50?  Well, it’s easy enough to calculate directly for small numbers, for any ratio.  As above, they all have an equal chance, so it’s only how many ways they can be arranged without changing the layout of heads and tails.  If you put numbers on each of them, you obviously get the factorial, but then you can rearrange the heads and tails without affecting the arrangement without the numbers.  So it’s the factorial of the whole divided by the product of the factorials of the number of heads and number of tails.  So if we’re going to 50/50 (assuming an even number of flips), the number of possible combinations is the factorial of the whole over the square of the factorial of half (for ten flips, 252).  Divide by the appropriate power of two.  So what happens to this as the number of flips gets too large to work it out exactly?

So let’s go back to that relation Wallis discovered.  Well, we’ll consider what happens when you actually write out that exact formula.  Up top, you get the factorial of the total number of flips, so the number of flips by one less and so on all the way down.  Then on the bottom you’ve got the square of the factorial of half that by the appropriate power of two, but there a funny thing happens - each factorial has half as many terms as there are twos, so sprinkling the twos in, you get the square of the product of just the even numbers up to the number of flips.  Cancel the even numbers out of the fraction, and you get the odd numbers over the even numbers.

Wait, I think I just explained the wrong thing.  The relation Wallis discovered, let’s consider what happens when you expand (1-x^2) to the power of some integer.  Well, it’s basically Pascal’s triangle, since you get the part that’s multiplied by one, and add it to the part that’s multiplied by x^2, which is one over.  It turns out that Pascal’s triangle is equal to the numbers we’ve been discussing for arrangements of coins, or as they’re also called, binomial coefficients.  Consider that for one flip, there’s one way to get no heads and one way to get one head, so that’s 1, 1, which is a row of the triangle.  Now just add up the fractions in the formula above, and you’ll see that the ways of getting a given number of heads in a given number of flips is the sum of the ways of getting the same number in one fewer flips and one fewer in one fewer.  So that’s Pascal’s triangle.  So that’s also what you get when you expand the integer powers of (1-x^2), i.e., (1 - 2*x^2 + x^4), (1 - 3*x^2 + 3*x^4 - x^6) and so on. And then you can use Cavalieri’s quadrature formula, that the area under the curve x^n is the difference between the values of x^(n+1)/(n+1) evaluated at the two points.  So for even n, between 1 and negative one, what you have is the sum of twice the coefficients divided by one greater than the power, e.g., for the formulae above, 2-2/3 = 4/3, 2-4/3+2/5 = 16/15, 2-6/3+6/5-2/7 = 32/35.

Now, with integration by parts, the chain rule, and the power rule (which is the same as Cavalieri’s quadrature formula only extending beyond integers), not only is it fairly easy to get this result, but it’s a more rigorous result than Wallis got.  He didn’t really prove what he claimed, but I’ve baldly asserted there that it can be proven with things vaguely resembling those mathy words.  I think you understand, human, but Wallis didn’t; regardless, where the integers were concerned, he recognized a pattern, that the integral of (1-x^2)^n is (2n)/(2n+1) times that of (1-x^2)^(n-1).

So the modern way to do that, again, would be by parts.  In effect, you would differentiate x*(1-x^2)^n and note that it's (2n+1)*(1-x^2)^n - 2n*(1-x^2)^(n-1), and note that x*(1-x^2)^n, provided n is strictly positive, is going to be the same at 1 and -1, so the integrals over that interval of the two terms of the derivatives must cancel each other out.  Wallis, who didn't have the product rule, couldn't have done that precisely (and I don't have it in me to plough through enough Latin to see how he actually did it), but again, beautiful concordance, thought itself - work out the right hand side term by term, and for each coefficient of x^(2k), k = 0 to n-1, you get (-1)^k*(2k+1) times the binomial coefficient, and for k = n, only the first term has such a coefficient, so you get (-1)^k*(2n+1), which takes the same form since n!/(k!*(n-k)!) = n!/(n!*0!) = 1.  Integrate, and the (2k+1) disappears from each term, so 2*(-1)^k times the binomial coefficient.  If n is even, this obviously cancels; if n is odd, you can see this cancels by the pattern in Pascal's triangle and the associative property of addition.

So anyway, this led to what’s known as the Wallis product.  Start with 2 - because obviously the area under a line of (1-x^2)^0 = 1 from 1 to -1 is 2 - and keep going to some extreme value, couple million maybe to warm up the old abacus, by using this formula.  So you'll want to multiply by 2/3, 4/5, 6/7, etc.  Then when you've got that massive tiny value, you shift half a step, and come back down.  Since you're half a step off, the numbers you would multiply by would be 3/4, 5/6, 7/8, etc., but since you're coming back down, it's actually 4/3, 6/5, 8/7, etc.  So starting with 2, to get to the area of the semicircle, you multiply by 2/3, 4/3, 4/5, 6/5, 6/7, etc., and then some error factor that goes to zero as n gets large because (2n)/(2n+1) approaches 1 as n gets large.

So let's take a tight, quartic uey back to ringin' a bell, John Wally be good, wasn't I saying something about aliens?  Anyway.  Remember we were at the product of the odds over the product of the evens.  But let's look at that product up there - 2 * 2/3 * 4/3 * 4/5 * 6/5 - even in the first few terms, it's clear that it's the square of the evens over the square of the odds, with a caveat - the odds will always be one further along than the evens.  So flip it over, stick a new even on the bottom, and take the square root, and you're going to the same place.  Since the product generated by the integrals goes to pi/2, the area of a semicircle of radius 1, the product generated by the coins goes to sqrt(2/(pi*n)), where n is the total number of flips, since remember that was the highest even.  So that's the approximate chance of getting exactly half heads for a very large even number of flips.

Okay, that incorporates the square root of pi, but, the humans ask, how does that come to a bell?  Well, consider a number of heads not especially close to all or none.  What's the quotient of that happening and of exactly half?  Well, consider what happens when you divide the binomial coefficients, you get increasing from the halfway point in the denominator, decreasing therefrom in the numerator.  The powers of two cancel, since it's still the same number of flips, so that's what you get.  So, for two thousand flips, let's pretend we had, say (for 970 or 1030 heads), (1000*999*...*971)/(1030*1029*...*1001), and reframe it as (971/1030)*(972/1029)*...*(1000/1001).  So consider the natural logarithm, i.e. log base e, pretending we're zoomed back far enough that the integral works as an approximation of the sum that arises from the log of the product.  We'll want to express it so that there's a one-unit gap between the log of each fraction.  So looking at it in terms of l, in this case thirty, you'll want to be integrating the natural log of the difference over the sum, or the natural log of (1-2t/(n/2+t)) from 0 to l, and to that end use the identity that the natural log of (1-x) is close to -x for very low x, which follows (switch the sign and take the log) from the definition of e, expressed as (1+x)^(1/x) for small x (switch the sign and take the log).  So we're looking for the integral of -2t/(n/2+t) = 2n/(n+2t) - 2, which is n*ln(1+2l/n)) - 2l.  (Consider the fundamental properties of the hyperbola, blah blah blah.)  The same approximation as before would make this zero, but that means it's off by infinity percent (my old math professor just slapped me), so instead use that fuckin' hyperbola to express ln(1-x) as the negative of the reciprocal of one minus t from 0 to t, and consider that you get this from a power series, e.g., one plus one third plus one ninth plus one twenty-seventh and so on will get you to three halves.  Integrate that, plug in -x rather than x, and you get that ln(1+x) - x is close to -x^2/2, so the approximation we're using is -2l^2/n.  That'll get you in this case .40657, as opposed to the true result of .40670.

Put it all together, and you get that for large l much smaller than n, the chance is approximately sqrt(2/(pi*n))*e^(-2l^2/n) (in the example above, 0.72537% as opposed to the true value of 0.72551%); this is as far as de Moivre got.  Already you've got something that, for large n, looks quite a bit like a bell, but it also looks quite a bit like a straight line.  For that reason, multiply the whole thing by the square root of n to keep it from vanishing, and to keep it the same area, multiply l by the same.  (So the ranges you'll be considering are in proportion to the square root of the total, which you already knew.)  What you get then is sqrt(2/pi)*e^(-2l^2), which is a normal distribution with a mean of 0 and a standard deviation of one half.  Why this isn't the standard normal is another story, to be told another time.

I point out to the now extremely bored aliens that this also means that the integral of e^(-x^2) over the whole number line must be the square root of pi, something that might be more interesting worked out backwards.  So consider a family of functions with discontinuities at integers, differentiable elsewhere with a derivative of zero, such that the value at every point is the binomial coefficient of twice the function's ordinal - call the ordinal n - and the function's ordinal plus the next natural up from the absolute value, being zero once x > n.  The integral of this on the whole number line will always be 2^(2n); redefine the function family by dividing by that and the integral will always be one.  Now, obviously this isn't approaching a function whose y-intercept is 1!  So multiply the whole thing by sqrt(pi*n/2) and change the discontinuities from the integers to the integers divided by sqrt(n/2); this will change the integral from 1 to sqrt(pi), and the tails will still vanish as n gets large.

Now let's look at the Riemann sums (defined for the sake of this proof as the sum of the values at the parameter closer to the origin times the length of the intervals) of e^(-x^2) from the negative square root of (pi*n) to the positive.  As n increases, this will approach the improper integral by definition.  Now, you've probably already sketched out the proof in your head and are wondering what's taking me so long.  The answer is that I'm a drunk who flunked out of college and now rambles on about aliens who can't do geometry.  So what's to be proven is that for every positive epsilon, there's both an n and a number of intervals to the Riemann sum such that increasing either will give a sum whose absolute difference from the nth function from the family above, plus the tails, is below epsilon.  We'll go with an n that's double the number of intervals, so that the approximation can be considered as the integral of a function with the same discontinuities as function n from the family, although the values won't be the same.  The value on either side of the y-axis would approach 1, as above, to a factor of at most 2n/(2n+1).  Because the rest of the limits were found by bounding the proportions to this value, this applies to all others.

As for the other limits, the argument from the integral clearly isn't enough at the edges.  So we have to put bounds on the edges as a whole as well as a bound in the middle, such that they all shrink.  Well, you just take the series expansion of the natural log and then that of the fraction and fuck it you're aliens you can take it from here.

Question:

The height of men (in inches) is the US is approximately normally distributed with mean \( 68  inches \) and standard deviation of \( 5  inches \).

1. A door frame manufacturer wants to build door frames such that only 2% of all men have to duck their head when walking through the door. How high does the door frame need to be?

2. The weight of men has a mean of \( 168  pounds \) and a standard deviation of \( 25  pounds \). The correlation between weight and height is 0.35. What is you best guess of a man's height if you were told that he weighs 160 pounds?

Solution:

Brief notes on Normal Distribution

Normal Distribution is a parametric distribution denoted \( N(\mu, \sigma) \) where \( \mu \) denotes the mean and \( \sigma \) denotes the standard deviation for the distribution. These two parameters completely describe the distribution. Normal distribution has a bell shape where you can easily find the percentage of possible values based on the distance from the mean you are looking at. This distance from the mean is usually measured in terms of number of standard deviations.

For example, these are some common measures

1. 68% of the possible values fall within 1 standard deviation from the mean i.e 68% fall in the range \( [\mu - \sigma, \mu + \sigma] \)

2. 95% of the possible values fall within 2 standard deviations from the mean i.e 95% fall in the range \( [\mu-2\sigma, \mu+2\sigma] \)

Similarly given a % of data points in question, we can find what range within the mean they all fall in. I will write more about Normal Distribution in a different post.

Door Frame Height

Let \( H \) be the required height for the door frames.

Average height ( \( \mu \) ) = 68 inches

Standard deviation ( \( \sigma \) ) = 5 inches

Converting \( H \) into a standard normal variable

\( z = \frac{H  - \mu}{\sigma} \)

Calculating the z-score based on the probability 

\( P(Z \le z) = 0.98 \implies z = 2.05 \)

Calculating the required height for door frames

\( H = \mu + 2.05\sigma = 68 + 2.05*5 = 78.1  inches \)

Correlation between height and weight

Height parameters: \( \mu_{h} = 68  inches \) and \( \sigma_{h} = 5  inches \)

Weight parameters: \( \mu_{w} = 168  pounds  \) and \( \sigma_{w} = 25  pounds \)

Correlation: \( \rho = 0.35 \)

Let \( H \) and \( W \) define the random variables for height and weight respectively.

\( H \) and \( W \) have a correlation of \( \rho \implies E[(H - \mu_{h})(W - \mu_{w})] = \rho\sigma_{h}\sigma_{w} \)

Convert the two random variables into standard normal variables for easier calculations. We can convert them back to their original values once we are done with the estimations.

\( z_{h} = \frac{H - \mu_{h}}{\sigma_{h}} \)

\( z_{w} = \frac{W - \mu_{w}}{\sigma_{w}} \)

\( Cov(z_{h}, z_{w}) = E[z_{h}z_{w}] = E[\frac{H-\mu_{h}}{\sigma_h}\frac{W-\mu_{w}}{\sigma_{w}}] = \frac{1}{\sigma_{h}\sigma_{w}}E[(H-\mu_{h})(W-\mu_{w}] = \rho \)

Consider the following scenario:

\( z_{h1} = \rho z_{w} \) and \( z_{h2} = z_{h} - z_{h1} \)

\( Cov(z_{h2}, z_{w}) = E[(z_{h}-z_{h1})z_{w}] = E[z_{h}z_{w}] - E[z_{h1}z_{w}] = \rho - \rho E[z_{w}^2] = 0 \)

This means that \( z_{h2} \) and \( z_{w} \) are independent. We have

\( z_{h} = z_{h1} + z_{h2} \)

\( E[z_{h} | z_{w}] = E[z_{h1} | z_{w}] + E[z_{h2} | z_{w}] = \rho E[z_{w} | z_{w}] + E[z_{h2}] = \rho z_{w} \)

Now that we have our general formula, lets plug in the numbers

\( z_{w} = \frac{160 - 168}{25} = -0.32 \)

\( E[z_{h} | (z_{w}=-0.32)] = -0.32\rho = -0.32*0.35 = -0.112 \)

Converting back to original variables

\( \frac{H-\mu_{h}}{\sigma_{h}} = -0.112 \)

\( \implies \frac{H-68}{5} = -0.112 \)

\( \implies H = 68 - 5*0.112 = 67.44  inches \)