[I’m writing this note while sitting at my son’s archery class. This is more of an attempt to use some “new to me” results to reconcile a puzzle I’ve been thinking about for 20 years than it is to write up something perfectly.]
In the mid 1990’s Zvi Bodie published a fascinating study on the price of long dated put options. The specific question he asked was about the cost of put that would be struck at a price that would guarantee an investment would at least achieve a return equal to the risk free rate (assuming 20% volatility). The paper is here:
Zvi Bodie’s “On the risk of stocks in the long run”
and here’s the abstract:
The answer to the question was surprising -> the cost of the insurance increased with the length of time:
An additional surprise is that the cost of the put increases to the value of the investment – i.e. it costs roughly $100 to insure that a $100 investment will at least grow at the risk free rate for 1,000 years.
The paper caught my attention in the 1990s because the math was so simple and the result seemed completely non-intuitive.
An old (but new to me!) talk by Ole Peters about the difference between time averages and ensemble averages helps explain the strange result:
The solution is the ensemble average growth rate will be equal to the risk free rate, r, but the time average growth rate will be the geometric growth rate . That difference means that as time grows larger the time average growth rate will fall farther and farther below the ensemble average – just as in the example in Peters’s video. Eventually (as time goes to infinity) the difference will be so large that the time average is essentially 0.
Saying that a bit more mathematically, as the expression
Thinking of the cost of the put more an insurer would think of an insurance premium also reconciles with the ideas from Peters’s lecture. Assuming a $100 initial investment, the present value of the maximum payout is $100 (since you are insuring risk free growth) . The only way the cost of the put can grow to $100 is that if as time grows you have a near 100% chance of a near 100% loss. Peters’s lecture shows how that is exactly what is happening with the time average.
Anyway, happy to have run across Peters’s work. It has given me a way to understand Bodie’s paper after pondering the odd result for 20 years.
My son had a little trouble with this problem from the 2009 AMC 10 B yesterday:
I should him how to solve the problem a few different ways – including by folding!
One of the ways I talked about was finding a rough answer by approximating by the fraction 9/4. We found this number by looking at a calculator and seeing that
As I thought about the problem more last night, I realized that 9/4 is part of the continued fraction approximation for . The first couple of approximations that you find using continued fractions are:
2, 9/4, 38/17, and 161/72.
If I approximate the hypotenuse of the original right triangle with those numbers, I get the following approximations for the length of BD, which are all ratios of consecutive Fibonacci numbers:
If you are familiar with continued fractions and especially the continued fraction approximations for the golden ratio, the emergence of the Fibonacci numbers probably isn’t a huge surprise. I missed it the first time, though, and think that students might really enjoy seeing this little Fibonacci surprise.
Today we revisited one of my all time favorite math projects for kids (also **revisiting**) :
We did this project once before, but I don’t think the kids remembered it:
An absolutely mind blowing project from James Tanton
The project is relatively simple to set up – you have strips of paper and make 5 Möbius strip-like shapes. If the short descriptions below aren’t clear, don’t worry, the videos have the “picture is worth 1,000 words descriptions
(1) An actual Möbius strip
(2) Same set up as making one Möbius strip, but you start with two strips of paper stacked on top of each other,
(3) A cylinder with a long oval cut out and a half twist on one of the strips left over after removing the oval.
(4) A cylinder with a long oval cut out and a half twist (in the same direction) in both of the strips left over after removing the oval.
(5) Same as (4) but the twists are in opposite directions.
Then you cut the shapes. In (1) and (2) you cut completely along the center line. In (3), (4), and (5) you cut around the oval.
What shape are you left with after the cutting?
(1) Cutting a Möbius strip
(2) Cutting two strips of paper folded into a Möbius strip
(3) Cutting a cylinder with an oval removed with one half twist
(4) Cutting a cylinder with an oval removed with two half twists in the same direction
(5) Cutting a cylinder with an oval removed with two half twists in opposite directions
Today’s project comes from this fantastic book:
The problem shows a neat connection between number theory and geometry -> what is the average number of ways to write an integer as the sum of (exactly) two squares?
We’ve looked this problem previously, but it was so long ago that I’m pretty sure that they boys didn’t remember it:
A really neat problem that Gauss Solved
I started by introducing the problem and then having the boys check the number of ways to write some small integers as the sum of two squares:
In the last video we found that the number 3 couldn’t be written as the sum of two squares. I asked the boys to find some others and they found 11 and 6. My older son then conjectured that numbers of the form couldn’t be written as the sum of two squares. We explored that conjecture.
My son’s conjecture was such an interesting idea that I decided to take a little detour and explore squares mod 4.
Slightly unluckily we were time constrained this morning, so the diversion in the last part left me with a tough choice about how to proceed. I decided to show them a sketch of Gauss’s proof fairly quickly. Don’t know if that was the right decision, but they did find the ideas and result to be amazing!
Even though I had to rush at the end, I’m really happy with how this project went. It is fun to see kids making number theory conjectures! It is also really fun to see gets really excited about amazing results in math!
If you’d like to see another fun (and similar) connection between number theory and geometry, Grant Sanderson did an amazing video about pi and primes:
Here are our the two projects that we did based on that video.
Sharing Grant Sanderson’s Pi and Primes video with kids part 1
Sharing Grant Sanderson’s Pi and Primes video with kids part 2
A few weeks ago I saw this tweet from the Joint Math Meetings:
I had to have it!
The stickers came today:
So, we found an old cube and replaced the old stickers with the new ones – here’s the final product:
If you’d like to order some for yourself, here’s the video from the folks who designed the product:
Saw a neat tweet from David Wees today. We are running around a bit tonight, but I had enough time to have the boys take a quick look before running out the door.
Here’s the problem:
Here’s my younger son talking through the problem:
and here’s what my older son had to say:
I saw in a later tweet that Wees intended to use this problem with younger kids (2nd and 3rd graders), but I’m still happy that the boys were able to talk through the problem and explain their reasoning.
Our friend Paula posted a video showing how to fold a small square out of paper. I thought it would be fun to ask my younger son to fold a small square before watching Paula’s video to see how he’d do it. Here’s what he did:
Next I asked him to watch Paula’s video:
Now I asked him to recreate Paula’s technique and talk through why it worked: