i960 

There's some parts of this analysis that while mathematically may indicate one thing, intuitively it doesn't totally add up for me.

Could you elaborate? I'm not sure whether your saying something about the analysis or about yourself.

Great post @SMCJB totally agree.

The opposite is only possible when only cash rich and a different mindset. I remember few points while a few years ago reading West of Wall St...the process was detailed how few pros avg down...but it calls for strict discipline and ability to take a loss if not working. Seen and done that...hence totally agree to your nugget.

Thnx
S

SMCJB 

Could you elaborate? I'm not sure whether your saying something about the analysis or about yourself.

I'm saying that while the analysis and math is good, there may be some practical/pragmatic/empirical points that don't necessarily match. For instance, the specific part about are stops being common or independent. After rereading your scenarios it does appear you are talking about common stops in the scale in/average down case and all in cases. However, are you talking about the same scenario with scaling in / averaging up (adding to winners)? Meaning the stops used for all positions are the same, hence in the adverse case of adding to the position and having it go against you the loss is the original position + the entire reversal portion of the move for the added size. If the analysis is assuming that the added to position has it's own stop then that's a different thing entirely.

In addition to all of this, this analysis appears as if p is static throughout the trade and set on entry. In reality, people averaging down/up/whatever are doing it at discrete areas where p is very likely to differ at different points throughout the trade. Would you agree that p at a support or resistance level probably differs compared to the middle of a range?

In fact, one might make the argument that just because price is going up doesn't mean it's going to continue going up - and that the risk of it going down grows as it keeps going up.

By no means would I argue for a martingale/losers average losers approach, I'm simply saying that traders may "blend" entries via scaling in because it's basically impossible to know the perfect entry point and as a side-effect of this they reduce immediate drawdown in the losing case. Most people using these techniques are doing them in an intelligent manner at specific areas of the trade and are actually trying to take advantage of the trade going against them. In many cases where said adverse move on entry doesn't happen the same traders will also scale in by averaging up.

i960 

There's some parts of this analysis that while mathematically may indicate one thing, intuitively it doesn't totally add up for me.

For your benefit, what doesn't add up is that the analysis conflates joint and conditional probabilities. The 3 scenarios have exactly the same expected payoff.









There's two simple ways to understand the mathematical intuition above:

1. Gambler's fallacy (https://en.wikipedia.org/wiki/Gambler%27s_fallacy)

2. Is it better to continue buying lottery tickets given that you have won a lottery?

Hope this helps.

artemiso 

For your benefit, what doesn't add up is that the analysis conflates joint and conditional probabilities.

With the greatest respect (and be assured i have a lot of respect for you) but I'm not sure I agree.

Firstly your probability diagram notation is a little confusing to me as the sum of your probabilities at time T+2 appears to be 3

To reiterate my base assumption

Quoting 

Ron Schoenberg and Al Corwin recently did some interesting research on the trading technique of "averaging-in". For e.g.: Let's say you have $4 to invest. If a future's price recently drops to $2, though you expect it to eventually revert to $3. Should you A) buy 1 contract at $2, and wait for the price to possibly drop to $1 and then buy 2 more contracts (i.e. averaging-in); or...

and maybe most importantly

Quoting 

In a moment of boredom I decided to expand Chan’s analysis to a binomial tree where we have a probability p of going down $1, but also a probability (1-p) of going up $1.

Hence p can range from 1 to 0 and when p = 1, there is a 100% chance we go down, but a 0% chance that we go up, and when p = 0, there is a 0% chance we go down, but a 100% chance that we go up.

Now take your example A.



When p = 1 (ie 100% we go down), both methodologies yield -4, which makes sense.
- (4-8) vs (-4-2+2)

When p = 0.5 (ie 50/50, both methodologies yield 0, which also makes sense.
- (4-4) vs (-1-1+2)

(note that between 1 and 0.5 they actually yield different results).

Now consider the case when p = 0. ie the market is 100% guaranteed to go up.
Your methodology yields 4 (4-0) but my methodology yields 2 (0+0+2).
If the market is 100% guaranteed to go straight up, there's never a chance to scale in and expected profit is 2 surely?

I understand that Bayes Theorem breaks down when p=0 or p=1 but since I don't concede that this is conditional probability I don't concede that Bayes Theorem applies here.

artemiso 

1. Gambler's fallacy (https://en.wikipedia.org/wiki/Gambler%27s_fallacy)

I understand conditional probability/the Monty Hall Problem etc etc.
I understand that if we flip a coin and it is a tail, that the probability that we then flip a second coin and it is a head is 0.5 not 0.25.
BUT that's not what we are saying here, we are saying whats the probability flipping two coins and getting a tail AND THEN a head in that specific order.

If you want to argue that the probability of flipping two coins and getting a tail and then a head in that specific order is 0.5 and not 0.25 then we can arrange to meet somewhere and I will bring a very large suitcase of cash with me.

In my analysis I handled the path dependency of this problem by treating a final outcome of 2 not as one outcome with a probability of 2*p*(1-p) but as two different scenario's, each with probability p*(1-p) that have different payouts. Hence maintaining the probability of the market ending at 2 to still be 2*p*(1-p)

I believe the key here is that we are not making the decision at time T+1 based upon what happened already, we are making that decision at T=0 BEFORE we know what happens. I do agree that if they were independent decisions taken at different times it would be different. That is not the case though, the decision to scale in was made before we enter the first trade. If it wasn't we would have bought more than we did.

artemiso 

2. Is it better to continue buying lottery tickets given that you have won a lottery?

That obviously depends upon the odds of winning.
If the probablity of winning again is 100% (ie p=0) then definitely yes.
As it is both are methodologies support continuing to buy lottery tickets when (p<0.5) as there is a positive expectancy.

Quoting 

Firstly your probability diagram notation is a little confusing to me as the sum of your probabilities at time T+2 appears to be 3

If you sum the joint probabilities at time T+2, you have (1-p)^2 + p(1-p) + p(1-p) + p^2 = 1.

SMCJB 

I understand that Bayes Theorem breaks down when p=0 or p=1 but since I don't concede that this is conditional probability I don't concede that Bayes Theorem applies here.

I hope you are aware that your method assumes that Bayes's theorem applies as well. The edges of a binomial tree represent conditional probabilities.

Here's an example on a peer-reviewed journal article using the binomial tree model: https://pages.stern.nyu.edu/~msubrahm/papers/HSS.pdf (see page 1132 and 1140 in the journal i.e. page 9 and 17 on the document) which should point out the conditional probabilities to you.

Quoting 

Now consider the case when p = 0. ie the market is 100% guaranteed to go up. Your methodology yields 4 (4-0) but my methodology yields 2 (0+0+2).

The transition probabilities are time-variant in this case, so you should really be distinguishing between p and p' and saying that p'=0 and P(W1)=1-p'=1.

Here, I quickly scribbled down what the probability tree will look like with initial P(W1) = 1 i.e. 100% guaranteed to go up at initial time step, then { p | p=/=0 and p=/=1 } of going down in the next time step:



In the second time step, as p -> 0, your expected payoff approaches +2.

artemiso 

If you sum the joint probabilities at time T+2, you have (1-p)^2 + p(1-p) + p(1-p) + p^2 = 1.

Agree completely.
As previously stated I think that properly separating the Up/Down and Down/Up to each have probability p(1-p) rather than saying the probability of the end node of 2 is 2p(1-p) address's the conditionality of the scale.

Specifically in case A our expected profit is
@4 ~ (4-2) * (1-p)^2
@2A ~ 0 * p(1-p)
@2B ~ 2*(2-1) * p(1-p)
@ 0 ~ [(0-2)+2*(0-1)] * p^2
which added together yields
-4p^2 -2p +2

artemiso 

I hope you are aware that your method assumes that Bayes's theorem applies as well. The edges of a binomial tree represent conditional probabilities. Here's an example on a peer-reviewed journal article using the binomial tree model: https://pages.stern.nyu.edu/~msubrahm/papers/HSS.pdf (see page 1132 and 1140 in the journal i.e. page 9 and 17 on the document) which should point out the conditional probabilities to you.

I'm 20+ years removed from my schooling and text books. I view myself as a trader with a very strong math background as opposed to a quant. As such I must concede that Theorem's & Lemma's etc are things in my past, and no longer second nature to me, as such I am unable to effectively debate with you in this manner. I will say though that the paper you reference appears to be addressing multivariate model as opposed to the univariate we are discussing.

SMCJB 

Your methodology yields 4 (4-0) but my methodology yields 2 (0+0+2). If the market is 100% guaranteed to go straight up, there's never a chance to scale in and expected profit is 2 surely?

artemiso 

In the second time step, as p -> 0, your expected payoff approaches +2.

So what am I missing? You seem to be supporting my case?

SMCJB 

So what am I missing? You seem to be supporting my case?

I'm not supporting your case.

I'm saying 3 things:

1. You're incorrect to say that Baye's theorem doesn't apply. Bayes's theorem does apply and you're even using it in your (misapplied) binomial tree. When you compute the joint probability in your binomial tree, e.g. p*p, p(1-p), (1-p)^2, why are these the product of 2 terms? Are you aware that 1 in each of these 2 terms is a conditional probability?

2. My model handles any p'=0 and p'=1 cases correctly. All that the p=0 and p=1 cases are saying is that there exist no outedges from that node in the probability tree. I assumed that this is not the case in my first post when I derived the expected payoffs, which is why you cannot plug in p=0 or p=1 into it. In my second post, I showed you that the result is still accurate if you introduce nodes with 1 outedge, e.g. initial step probability p'=0 and second step probability 0<p<1.

3. This is the easiest part to see where you went wrong:

Quoting 

Case iii) the market drops at T+1, and then rallies back to $2 at T+2. In that case our initial contract purchased at $2 is breakeven, but when prices dropped to $1 at T+1 we added two more contracts. These two contracts now have a total profit of $2. Hence the expected payout of case iii) is 2*p*(1-p). Finally in case iv) the prices drops at T+1 and T+2 and ends at $0. In this case we lose $2 on our initial one contract purchase and lose an additional $2 on our two extra purchases at $1. Hence we have a total loss of $4 and an expected payout of -4*p*p

P(W1|W2) denotes the probability of the "if W1 then W2" event. This is called a conditional probability.

P(W1 and W2) denotes the probability of the "W1 AND W2" event. This is called a joint probability.

The 'If market drops at T+1 then market drops at T+2' event has probability p.

The 'Market drops at T+1 AND market drops at T+2' event has probability p^2.

Where you're erroneously getting the quadratic terms in your calculation is that you're multiplying the (W1|W2) outcome value with the probability of the (W1 AND W2) event, which is meaningless.

Another way to look at this: Suppose I took a 6 chamber revolver with 1 bullet and played 100 rounds of Russian roulette unharmed. Is it better to use this revolver (add to my 'winners') than a completely identical 6 chamber revolver with initially2 bullets that has fired off once (now has 1 bullet)?

SMCJB 

With the greatest respect (and be assured i have a lot of respect for you) but I'm not sure I agree.

Well, that's why I deleted my post in page 1. I don't mean to create any ill will between us. I figured you'd see my deleted post in your email subscription and correct the error yourself, or perhaps someone else would do it.

Quoting 

there exist no outedges from that node in the probability tree

I meant there exists only one outedge from that node in the probability tree...