Impossible to find a real pro with a real edge as a mentor?

Why would a well heeled trader (Read a winner) want to mentor anyone unless he/she felt a DESIRE to do such a task? I mean come on, place yourself in their shoes and ask yourself a simple question? Do i really want to do this? I do not need the money or the recognition from this person unless i felt a deep desire to be of service for such a task.

That boils down to a very, very selective group because the work is more than the reward. Follow? A winning trader can make a heck of a lot more money by doing what he/she has been doing up to the point where even the thought of mentoring is like asking for a JOB. GOOD GRIEF Charlie Brown!!!!

i have said before and will still say: I might someday bring in a single or two young teens as a summer job and let them be "TRADE PARTNERS" with me. Yes, i am a democrat and believe in a fair days pay for a fair days wages, ha. If i traded X amount of cars i would put up the same amount of cars for each of their trades. They keep the profits and get to swallow the losses same as me. How is that They would in no time want to be calling some trades for all of us. Their parents name would be on the accounts and naturally i would be the one trading the account for their parents. That sounds legal anyway....(you get the picture).

Think about it.........What better way to learn than being thrown into battle with bullets flying and many (MOST) days make more than their parents.

In my book mentoring is about "GIVING" not receiving. In my book there is a certain satisfaction about giving something back. Losers in life expect to receive more than they give.

goldenarm has some mighty fine posts in this thread, he is right on with succinct, direct words......MY kind of guy. Traders like myself sometimes come across as curt but after being a trader for many a year being curt comes with the territory because daytraders do not have time to be sweet and sugar coated when it comes to making decisions.

TODAY as example another sports person said what you will hear from many professionals in many a profession when you are damm good at what you do. Kobe Bryant ( i am not a big fan of basketball) said Gasol was not having a very good first half so at 1/2 time Kobe said to him: "You're thinking to much,............ STOP thinking so much and just go out there and play"

I will spare you the typing of how many times i have heard that from professionals.............trading a 5 minute chart is about TRADING, NOT thinking and pretending.
 
Quote from Loki:

If you are a real pro...with a real edge and you make real money, why would you want to be a mentor ?
More...
because after a while making money can become very boring,i'm the farthest thing from that point, but a friend was,he was burned out, and left to coach high school foootball,five years later he resumed making even bigger money
 
Well Merry Christmas to all.

My problem with "my" trading has been, not really learning from my losses. My losses and wins have all felt the same to me. I look at the screen and look for good buying opps but whether the trade goes in my favor or against me, nothing shoots at me saying something different.

I have a full time job but have always loved trading. I know I don't have a edge and trade for enjoyment and I am trying to learn but nothing seems to seep into my brain.

Not sure why I am posting but I do have some Christmas juice in my system now but if someone has a little advise to share I am all ears.

Everyone take care and hope you all have a great 09
 
Quote from shortie:

statistics does not come naturally to me :(.

i was not able to figure out the formula for probability ratios. i probably lack some basic knowledge in statistics.

Here is what i tried and got stuck.

X - stop
Y - outcome
P(<=X) - probability that the price moves to X (or lower) any time during experiment
Pend(>=Y) - probability that the price ends up at or above Y at the end of experiment
Pend(X_Y) - probability that the price ends up between X and Y at the end of experiment

we are interested to know the ratio P(<=X)/Pend(>=Y)

the above probabilities add up to 1:
P(<=X)+Pend(X_Y)+Pend(>=Y) = 1

but i don't see how to get to P(<=X)/Pend(>=Y) from the above.


without formulas i am guessing that the monkey loses money no matter what X. to make money it should remove stop and add a limit to lock in profits.
More...

Shortie:

First I want to encourage you for having taken the time to think about the problem. You will see the benefit later.

I would also like to ask why you insist on solving the MOST general case (any X and Y)? Just try to think about the "easier case" as explained below. The reason I asked for the ratio is that one can find it without know the probability. This will be interesting because it would mean that the result is valid for ANY distribution of stock prices.

Also a clarification: your X is a price, while in the statement of the problem X is meant to be the distance from the stock the stop price.

Let as add another variable. Let's call it z.

Stock at price S.

Stop at S-z (if long). Adopting you understanding of X, we have have X=S=z

Profit level is Y such as: Y/S=S/(S-z). If Z is small, Y~S+z.

So study the problem as a function of z. (Notice that you have one variable only).

If you cannot crack anything from it yet, I suggest you run a simulation in a spreadsheet.

However the answer will surprise you. I will prove the result in two lines!

I will also send you the answer in a PM to you as agreed (with a clear title so that you can read it only when you feel you want).


Cheers!

PS: Do you know if russia's market has american-style options traded?
 
RFT, I know you are talking about some basic but very important stuff. So, I want to make sure I get ito.

Here is the set-up:
Stock price is S, the monkey goes long @S in the morning. it puts Stop @(S-z). it closes the position EOD (or is stopped out).
we want to know the ratio of the probability that the stop is hit to the probability the stock closes EOD >=(S+z).

I have diagrammed that there are 4 intervals where the price could travel (it actually never gets to A because of Stop):

---A----||----B---||---C---||----D---
..........S-z...........S.........S+z..........

I can only think about this thing in terms of probabilities and the fact that X=Y does not help me. What confuses me is that
we have to talk about different probabilities here (two EOD probabilities and one about happenings any time during the day):
P(A) - probability the price will reach interval A ANY time during the day
Peod(B-C) - probability the price will reach intervals B or C EOD
Peod(D) - probability the price will reach interval D EOD

again:
P(A)+Peod(B-C)+Peod(D)=1

and we need P(A)/Peod(D)

the fact that B=C does not help because not all probabilities EOD and I can't see how they relate to each other. Peod(B-C) seems to be the main problem here.

I may try the spread sheet if i could set it up quickly. i probably should have a set-up to play with a random walk anyways.


i think the russian market has a daily volume ~$10M. i don't know about options but the total dollar volume makes me suspicious about the market itself.

Quote from riskfreetrading:

Shortie:

First I want to encourage you for having taken the time to think about the problem. You will see the benefit later.

I would also like to ask why you insist on solving the MOST general case (any X and Y)? Just try to think about the "easier case" as explained below. The reason I asked for the ratio is that one can find it without know the probability. This will be interesting because it would mean that the result is valid for ANY distribution of stock prices.

Also a clarification: your X is a price, while in the statement of the problem X is meant to be the distance from the stock the stop price.

Let as add another variable. Let's call it z.

Stock at price S.

Stop at S-z (if long). Adopting you understanding of X, we have have X=S=z

Profit level is Y such as: Y/S=S/(S-z). If Z is small, Y~S+z.

So study the problem as a function of z. (Notice that you have one variable only).

If you cannot crack anything from it yet, I suggest you run a simulation in a spreadsheet.

However the answer will surprise you. I will prove the result in two lines!

I will also send you the answer in a PM to you as agreed (with a clear title so that you can read it only when you feel you want).


Cheers!

PS: Do you know if russia's market has american-style options traded?
More...
 
What your just wrote is the clearest understanding of the problem I have read from your you (and obviously only you) so far

In addition, importan stuff I think so. Basic/easy, I am not sure about that. Everything looks easy in hindsight, but not necessarily in foresight particularily one does not even know the question to pose.


Two remarks:

1. This is what is causing you the problems: P(A)+Peod(B-C)+Peod(D)=1

The above is true ONLY if A is meant for EOD. If not EOD the "=1" is wrong.

2. Russian Market: If you have options, the more they rigg the equity market, the greater it is (for me!). Please check those options. I told you in PMs hints, but we cannot discuss such things in here.
 
Quote from riskfreetrading:

What your just wrote is the clearest understanding of the problem I have read from your you (and obviously only you) so far

In addition, importan stuff I think so. Basic/easy, I am not sure about that. Everything looks easy in hindsight, but not necessarily in foresight particularily one does not even know the question to pose.


Two remarks:

1. This is what is causing you the problems: P(A)+Peod(B-C)+Peod(D)=1

The above is true ONLY if A is meant for EOD. If not EOD the "=1" is wrong.

2. Russian Market: If you have options, the more they rigg the equity market, the greater it is (for me!). Please check those options. I told you in PMs hints, but we cannot discuss such things in here.
More...

Peod(B-C) and Peod(D) are meant to be probabilities when stop is implemented. I should have called them P'eod to indicate that they are different from the EOD probabilities without Stop.

with the above correction i can write:

Peod(A)+Peod(B-C)+Peod(D)=1

P(A)+P'eod(B-C)+P'eod(D)=1

we would like to know P(A)/P'eod(D)
 
Quote from shortie:

Peod(B-C) and Peod(D) are meant to be probabilities when stop is implemented. I should have called them P'eod to indicate that they are different from the EOD probabilities without Stop.

with the above correction i can write:

Peod(A)+Peod(B-C)+Peod(D)=1

P(A)+P'eod(B-C)+P'eod(D)=1

we would like to know P(A)/P'eod(D)
More...

Now you are making me doubt whether you have the correct understanding of the problem. ALL probs are meant to be probabilities when stop is implemented.

The difference between the values sought is in the nature of the events. Therefore I would denote the one with a dash as the ones that correspond to a trade which is stoped, and the ones without a dash the ones that correspond to EOD events.

Thererefore we are looking for: P'eod(A)/P(D).

I think that the way you want to solve the problem gets you to go back and slightly change the problem without you realizing it? (unless I mis-understand).

Instead of finding the ratio, and to get you out from the way you now seem to think about the problem, I will ask this:

could you compare P(A) to P'eod(A)?

This will put your mind to work on the difference between the P() and the P'eod() since the event (A) is the same for both.

I guess that this will get your wheel out from the spinning.
 
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