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Why you should add to winners and never add to losers


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Why you should add to winners and never add to losers

  #21 (permalink)
 
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 SMCJB 
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artemiso View Post
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.

Just to confirm there is no ill will on my side, hopefully none on yours either.

Unfortunately this is the ‘thread subscription notification’ email that I got.


And when I went to the actual post to see what the attachments were I got.


You actually made two posts that morning, but the way notifications work I didn’t get a notification for the second post, and hence have no idea at all what it said.

I know you delete a lot of your posts for privacy reasons, something I try to honor by not quoting your entire post, but since I didn’t & don’t think I was wrong, my assumption was – ready for a laugh – that you had deleted your post because you realized you had made an error. I was obviously very wrong on that!

Regarding your last line, “and correct the error yourself, or perhaps someone else would do it” I must say that with all the 1000s of views this post/thread has received and the >50 like’s, that if the analysis is off as you say, I’m surprised somebody else hasn’t pointed it out. Also surprised that nobody has weighed in on this follow up discussion (other than i960.)

Returning to the debate at hand, I believe I may be losing the war of words, but don’t believe I’m necessarily losing the war of facts. As I previously stated I have a lot of respect for you, and am under the impression that your knowledge of math and our area’s of finance, is superior to mine. This is all second nature to you, and you can talk very authoritatively and eloquently on the subject. As previously stated. I’m 20+ years away from my degree in math, and it’s no longer all second nature to me. I even had to check that Bayes Theorem was what I thought it was. While I believe the underlying message and intent of my replies to you are correct I acknowledge that I have miss-worded and ambiguously worded certain statements.

Unfortunately saying all of that I still disagree with you that the payout for all 3 scenario’s is (4-8p). Let me try and illustrate why once again, and hopefully you can explain to me in layman’s terms why I am wrong.

To reiterate the situation, at time T the current price is $2, at T+1 it can move up to $3 or down to $1, and at time T+2 it can move up/or down $1, yielding prices of $0, $2 or $4. At each time step the probability of a $1 drop is p and the probability of a $1 rise is (1-p).

In scenario C we do nothing at time T but if the price drops to $1 at time T+1 we will buy 4 contracts at $1. Note that this decision is being made at time T and not at T+1. We can’t/won’t change our mind.

It is my opinion that at time T, the expected payout of Scenario C is (4p-8p^2)

I believe it is your opinion that at time T the expected payout of Scenario C is (4-8p). (I believe that this is in fact represents the expected payout at Time T+1 given that we are trading $1 and not the expected payout at time T.)

The chart below illustrates the trajectory of these two payoffs at points (0, 0.01, 0.1, 0.2 … 0.9, 0.99, 1.0)




As p -> 1, ie the market is guaranteed to go down, both our payouts -> -4, which makes sense as we are long 4 contracts at $2 and the market is almost guaranteed to drop to $1.

As p -> 0, ie the market is guaranteed to go up, my payout -> 0. This makes sense to me as I have no position on
and the market has gone up before I can establish a position.

As p -> 0, your payout -> 4. How can this be possible if you don’t have a position on?

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  #22 (permalink)
 artemiso 
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Quoting 
I must say that with all the 1000s of views this post/thread has received and the >50 like’s, that if the analysis is off as you say, I’m surprised somebody else hasn’t pointed it out. Also surprised that nobody has weighed in on this follow up discussion (other than i960.)

I can just as easily find you a post with 50 likes on "R:R multiples". Which is why there is good money to be picked off from these guys.


Quoting 
In scenario C we do nothing at time T but if the price drops to $1 at time T+1 we will buy 4 contracts at $1. Note that this decision is being made at time T and not at T+1. We can’t/won’t change our mind.

It's not about changing your mind. You said it yourself: IF then price drops to $1, THEN we will buy. You're looking at P(we buy | price drops to $1), not P(price drops to $1 AND we buy).

Suppose if the sun doesn't rise at time T+1, I will let you play this game at time T+2: I pay you $6M if you roll a six on a fair dice and $0 otherwise. I let you decide at time T your preplanned action at time T+2 if the sun doesn't rise.

P( sun doesn't rise tomorrow) is approximately zero, so according to you, the expectancy of this is P( sun doesn't rise tomorrow AND you roll a six )*$6M -> $0. This is consistent with your suggestion not to 'add to losers', i.e. at time T, you would not to play this game at T+1.

However, the expectancy of this 'adding to your losers' is actually simply P( you roll a six | sun doesn't rise tomorrow)*$6M = 1/6*$6M = $1M. At time T, I can already preplan that if the sun does not rise at T+1, I will play this game at T+2.

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  #23 (permalink)
 artemiso 
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SMCJB View Post
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.

By the way, this is another place where your methodology is incorrect. I gave the example of lotteries deliberately because this is a binomial experiment.

Buying n tickets upfront (all-in at initial time step) gives you odds of n/N. Suppose probability of success p such that p+q=1. The odds of success from buying over n steps are:



which is worse than all-in upfront.

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  #24 (permalink)
 artemiso 
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artemiso View Post
By the way, this is another place where your methodology is incorrect. I gave the example of lotteries deliberately because this is a binomial experiment.

Buying n tickets upfront (all-in at initial time step) gives you odds of n/N. Suppose probability of success p such that p+q=1. The odds of success from buying over n steps are:



which is worse than all-in upfront.

P.S.: Big N is the number of combinations in the lottery.

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  #25 (permalink)
 artemiso 
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SMCJB View Post
As p -> 0, ie the market is guaranteed to go up, my payout -> 0. This makes sense to me as I have no position on
and the market has gone up before I can establish a position.

As p -> 0, your payout -> 4. How can this be possible if you don’t have a position on?

Exactly, this is a conditional probability!

Assume the current market price is $100M, and market is practically guaranteed to go up from $1 if it ever gets there, i.e. p ->0.

You can replace your strategy with 'If market price drops from $100M to $1 and Taylor Swift marries me, then I will buy 4 futures contracts @ $1.' The expected payoff of your strategy is still $4 because your strategy decides what to do now *if* Taylor Swift marries you in the future.

An assessment of how good this strategy is is independent of how likely Taylor Swift is going to become your wife.

That's why all 3 scenarios have equal expected payoff.

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  #26 (permalink)
 
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 SMCJB 
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While you've given some colorful analogies, Gamblers Fallacy, Lottery Tickets, Sun Rising Tomorrow, Russian Roulette, and even marrying Taylor Swift you still have not explained how you can make money when you don't have a position on?

Re: the lottery discussion. You said

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

To which I replied...

SMCJB
That obviously depends upon the odds of winning.
If the probablity of winning again is 100% (ie p=0) then definitely yes.

Which I obviously still stand by. If you have a 100% chance of winning who wouldn't play.
But then you say..

artemiso View Post
By the way, this is another place where your methodology is incorrect. I gave the example of lotteries deliberately because this is a binomial experiment.

Buying n tickets upfront (all-in at initial time step) gives you odds of n/N. Suppose probability of success p such that p+q=1. The odds of success from buying over n steps are:



which is worse than all-in upfront.

What you originally asked (buy, win, continue buying), and what your now saying (buy, win, continue buying) vs (buying tickets upfront) are two different scenario's and the second is not the scenario I answered. Either way I believe my response to be correct to the original question. If you believe that you shouldn't buy a lottery ticket when the chances of winning are 100% I'd be curious to hear why. (Assuming of course that the ticket price isn't more than the prize!)

Maybe we're debating different things and don't realize it.

artemiso
Suppose if the sun doesn't rise at time T+1, I will let you play this game at time T+2: I pay you $6M if you roll a six on a fair dice and $0 otherwise. I let you decide at time T your preplanned action at time T+2 if the sun doesn't rise.

P( sun doesn't rise tomorrow) is approximately zero, so according to you, the expectancy of this is P( sun doesn't rise tomorrow AND you roll a six )*$6M -> $0. This is consistent with your suggestion not to 'add to losers', i.e. at time T, you would not to play this game at T+1.

However, the expectancy of this 'adding to your losers' is actually simply P( you roll a six | sun doesn't rise tomorrow)*$6M = 1/6*$6M = $1M. At time T, I can already preplan that if the sun does not rise at T+1, I will play this game at T+2.

I agree that I am saying P( sun doesn't rise tomorrow AND you roll a six )*$6M -> $0.

I also agree that P( you roll a six | sun doesn't rise tomorrow)*$6M = 1/6*$6M = $1M ~ BUT that's NOT what I'm saying. I am saying P( sun doesn't rise tomorrow AND you roll a six )*$6M -> $0.

Note no emphasis added to the AND. That was your own wording.

In your example you say "Suppose if the sun doesn't rise". In my analysis we are evaluating scenarios A, B, and C "whether the sun does or doesn't rise" to use your expression and not "ONLY when it doesn't rise". I agree that the expectancy of adding to losers, at the time we add to them is what you say. That's not we're analyzing though.

Put another way, how about this. I offer you two opportunities, you pick one.

Opportunity One. "If the sun does rise tomorrow, I will let you play this game where I will pay you $6M if you roll a six on a fair dice and $0 otherwise.

Opportunity Two. "If the sun doesn't rise tomorrow, I will let you play this game where I will pay you $6M if you roll a six on a fair dice and $0 otherwise.

Are you really going to say that they have the same expected payout and your indiffierent to either oppurtunity?

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  #27 (permalink)
 
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Interesting - @SMCJB, you're implying that it's not necessarily a matter of statistical efficiency, but a pragmatic matter. Theoretical vs practical.

Aside from that, I've never had to think so hard reading a thread. Thanks to both of you guys

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  #28 (permalink)
 artemiso 
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SMCJB View Post
Opportunity One. "If the sun does rise tomorrow, I will let you play this game where I will pay you $6M if you roll a six on a fair dice and $0 otherwise.

Opportunity Two. "If the sun doesn't rise tomorrow, I will let you play this game where I will pay you $6M if you roll a six on a fair dice and $0 otherwise.

Are you really going to say that they have the same expected payout and your indiffierent to either oppurtunity?

This is not analogous to scenarios A, B or C. Do you see why?

In Scenarios A, B or C, you are making an action after the realization (market data) at time T+1. I gave the correct analogy to this: I said that you can precompute ahead of time that if the sun doesn't rise at T+1 that it yields an expectation of +$1M to play the game. So when the sun does indeed not rise at time T+1, then you would play the game. Your action time is T+1 and your decision space is {bet on rowing a 6, don't bet on rowing a 6}.

In the two opportunities you are giving me, you are only allowing me to make an action at time T (before the realization at time T+1) predicated on me already fixed on playing the game. In this case, the expectation values that you should be comparing are P( sun rises AND row a six | play the game)*6M and P( sun doesn't rise AND row a six | play the game)*6M. Your action time is T and your decision space is {bet on sun rising, bet on sun not rising}.

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 i960 
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One of the other issues in this debate is that it seems the extremes of 0% and 100% are being used to prove or disprove a point. As I said earlier, p is not necessarily static throughout the decision process. Perhaps what I meant by that was that they're independent probabilities occurring at disparate moments in time. They appear linked, but aren't.

If we want to get practical/pragmatic and look past the 0%/100% p stuff, when we consider things like S/R, ranging >80% of the time, yadda yadda, then technically adding to losers should actually outperform adding to winners over the long-term in the majority ranging case. Obviously in the hard core trending situation this won't be the case. I know this, when I add to a position that's gone against me, I'm not adding because I'm trying to dig out of a hole quicker, I'm adding because I still believe in it and still expect it to reverse. The initial position could be considered a "get in the trade" at a reasonable (but perhaps not perfect) entry point.

It also was never answered in the original post if adding to winners used the same stop as the original position.

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i960 View Post
One of the other issues in this debate is that it seems the extremes of 0% and 100% are being used to prove or disprove a point. As I said earlier, p is not necessarily static throughout the decision process. Perhaps what I meant by that was that they're independent probabilities occurring at disparate moments in time. They appear linked, but aren't.

While I agree that p = 0 or 1 are extremely unlikely cases, they are obviously (along with 0.5) the easiest cases to illustrate. Your also right that in real life "p is not necessarily static throughout the decision process" but if you can correctly identify auto correlation/serial correlation there are probably more efficient ways to take advantage of that, than by this. I was just trying to expand on Chan's analysis and illustrate why adding to losers is a bad idea.


i960 View Post
It also was never answered in the original post if adding to winners used the same stop as the original position.

I'm sorry but I thought it was obvious that this was just a simple 2 step analysis and that stops were not considered. Expanding the analysis and looking at stop rules, and how they effect the decision making would be an interesting exercise.

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