Progress Report 34
Bearish Wolf
I mentioned yesterday that I would be reading up on the book by Natenberg. Well, no, I decided I am not doing that. I am taking it as a reference, for sure, but I am taking a shortcut to learn option pricing: just reading up on lecture notes. I have been lucky enough to find lecture notes that summarize the Black-Scholes model and the greeks. And that is everything I need. I was to embark in a journey of reading books of 1.000 pages each but gladly I realized that I could could achieve my goals faster.
Time Crushing: Estimated time to read Natenberg is around 50-80 good hours. Estimated time to read the lecture notes and supplement that knowledge with Natenberg? About 5. So let’s be lazy/efficient and implement only the actions that actually make sense to bring us to our results. Let’s never do things just ‘to be busy’. So this was the focus of today: downloading those lecture notes and rushing through them. Around 40% done already.
Some other stuff I have been up to is mental Mathematics. I will eventually put up an entry concerning only these strategies. For a sneak peek of this, scroll to the end of this blog entry.
Moreover, I am committing to at least one trade per week. And the winner is the following one (as suggested by another trader):
- Monday or Tuesday if Monday is a holiday, between 3:30pm – 3:57 pm, sell to open (naked) 1 contract (100 shares worth) of UVXY ATM.
Besides this, I need to start thinking about resetting my paper trading account to mimic my real future portfolio. What I mean with this is that currently I have a paper trading account in IBKR with around $1.000.000 in buying power. My personal buying power is different than this, so I need to open another trading account that would be a ‘real-life’ simulation of what would happen with a portfolio of my size, with the trades that I do.
Following is a small draft of the blog regarding mental Mathematics that I will eventually put up in my blog. For now, ignore all examples, as they should be rendered in LaTeX, but I have not figured out how to do that just yet in HTML. For a better view of this, go to bullishwolf.wordpress.com
Multiplication
For notation purposes and explanation purpose I am going to use the terms
multiplicand and
multiplier. The one that is being ‘multiplied’ is the multiplicand and the one ‘multiplying’ is the multiplier. In low orders of magnitude, this does not really matter. e.g. in 7*6 it does not really matter what multiplies what, the answer is straightforwardly a 42. But in higher dimensions, such as with 34*2, it does matter what multiplies what, since you would want the multiplier to be the easiest one to multiply the other. Here, for example, we would one 2 to be the multiplier to 34 as it is quicklier to calculate 2 times 34 (34 + 34 = 68) than 32 times 4 (2 + 2 + 2 + … + 2 = 68). Of course, because of the commutative law, both answers yield the same results, but the former is easier in your head.
Use the 5’s to your advantage.
There is a very easy way to multiply any number by 5. It goes like this:
- Make five the multiplier.
- Take the multiplicand and multiply it by ten, i.e. just add a zero at the end.
- Divide this number by two. This is your final answer.
Since divisions by two are most of the times easy to carry out, this method saves you a lot of the trouble going through all the motions of the multiplication. Here is an example:
$$354 * 5 = 345*\frac{10}{2} = \frac{3540}{2} = 1770$$
Use the 11’s to your advantage
Every multiplication by eleven can be carried out in seconds. The trick is as follows:
- Make 11 the multiplier.
- Anchor the initial and the last number of the multipliand. These are also going to also be the initial and last numbers of your final answer. Sometimes you will have to add one to the initial anchor, but we will come to that later.
- Starting from the right of the multiplicand, sum each two individual consecutive numbers.
- If the sum is one digit, this is the next number of the answer next to the last one (starting from the right). If the sum is two digits, the last digit is the next number of the answer, and the initial digit you carry over to the next sum. Do this algorithmically until you get to the first digit. Couple this with the anchors in step two and you have the answer.
Sounds a little bit more complicated than what it actually is, so here is an example:
$$\text{1. }423*11 \ \text{2. }\text{Anchor the 4 and the 3 } \rightarrow 4…3 \ \text {3. }\text{Start summing each two consecutive number from right to left:} \ 3+2 = 5 \ 4+2 = 6 \ \text{4. } \text{Couple these sums with the anchors: } …56…\text{ & } 4…3 \rightarrow 4563 \text{ (final answer)}$$
A more complicated one:
$$\text{1. }6827*11 \ \text{2. }\text{Anchor the 6 and the 6 } \rightarrow 6…7 \ \text {3. }\text{Start summing each two consecutive number from right to left:} \ 7+2 = 9 \ 2+8 = 10 \ \textit{ keep the zero, carry over the 1} \ 6+8=14 (+1) = 15 \\textit{ keep the 5, carry over the 1} \textit{ to the initial anchor} \ \text{4. } \text{Couple these sums with the anchors: } \ …509…\text{ & } (6+1)…7 \rightarrow …509…\text{ & } 7…7 \rightarrow 75097 \text{ (final answer)}$$
Also works for simple two digits:
$23*11 = 253. $
Multiplying Numbers With Same Initial Digits and Last Digits Summing Up to Ten (2X2 multiplications).
Let use introduce a new operator, the concatenator $\operatorname{CC}{ x,y }$. This operator does exactly what you would expect: it just concatenates two numbers, one after the other. e.g. $\operatorname{CC}{ 12,25 } = 1225$. We will see shortly why this is helpful.
Now, for every multiplication of two digits by two digits, starting with the same number (e.g. 34*36), and for which the last two numbers sum up to ten (e.g. 4+6=10), there is a fairly straight forward way to compute the answer.
- Multiply the initial number by the the number right above it.
- Multiply the last digits by each other.
- If the result of two is double digit, concatenate the results of (1.) and (2.). If the result is double digit, concatenate the result in (1.) and (2.), but with a zero in between.
Here are two examples:
$$36
34\ \text{1. }\text{Multiply 3 by the number right above it:} \ 3(3+1) = 12 \ \text{2. Multiply the second numbers by each other:} \ 6*5 = 30 \ \text{3. } \text{Since the result of (2.) is two digits, we just concatenate these results together: } \ \operatorname{CC}{12,24}=1224 \text{ (final answer)}$$
$$51
59\ \text{1. }\text{Multiply 5 by the number right above it:} \ 5(5+1) = 30 \ \text{2. Multiply the second numbers by each other:} \ 1*9 = 9 \ \text{3. } \text{Since the result of (2.) is one digit, we add a zero in between} \ \text{and concatenate these results together: } \ \operatorname{CC}{30,0,9}=3009 \text{ (final answer)}$$
