Let's say that I want to buy a stock like MOMO and I want to buy it today (this decision was made before the market open). The duration of my position will be between a week and several month. When should I buy the stock so that I'm nearly sure that I won't lost money in that day (because of speculation).
The best time to buy a stock is the time of day when the stock price is lowest!
Obviously you learned nothing from that sentence, but unfortunately you won't get a much better answer than that. Here's a question that is very similar to yours: "Is it better to have a picnic for lunch or for dinner to minimize the chance of getting rained out?" Every day is different...
You can't predict when to buy a stock during the day to guarantee not having a loss for the day. In the short run stock prices are really pretty random. There are many day traders who try to accomplish exactly this and most of them lose money.
If you don't believe me, create an account on Investopedia and use their free stock market simulator and try day trading for a few months.
One of the biggest laws in economics is that if an opportunity is very profitable and is very easily exploitable even by complete beginners, then it will very soon stop being profitable.
That's how the market works. If you buy stock when it is at the lowest, then you are making money, but most of the time someone else is losing money. And if there was a magic hour of the day when buying would be the most profitable, then soon everybody would want to buy at that time and no one would want to sell anything, so the scheme would collapse.
The best thing to do is not worry about what time is best to buy but put in a conditional order before the market opens. If your conditions are met during the trading day your order will go through and you will buy the shares.
This keeps your emotions out of your trading and will stop you from either chasing the market or buying when you consider the wrong time.
As you have already done your analysis and made your decision before market open, thus you should place your conditional orders and stop losses before market opens as well.
You want to buy when the stock market is at an all-time low for that day. Unfortunately, you don't know the lowest time until the end of the day, and then you, uh can't buy the stock...
Now the stock market is not random, but for your case, we can say that effectively, it is.
So, when should you buy the stock to hopefully get the lowest price for the day?
You should wait for 37% of the day, and then buy when it is lower than it has been for all of that day.
Here is a quick example (with fake data):
10 9 7 5 8 12 14 23 14 8 14 3 9 11 1 12 3 12
We have 18 points, and 37% of 18 is close to 7. So we discard the first 7 points - and just remember the lowest of those 7.
10 9 7 5 8 12 14 23 14 8 14 3 9 11 1 12 3 12 ────────┬─────── Discard this 37%
We bear in mind that the lowest for the first 37% was
Now we wait until we find a stock which is lower than 5, and we buy at that point:
10 9 7 5 8 12 14 23 14 8 14 3 9 11 1 12 3 12 └───────┬──────┘ └──────┬─┘ | Discard this 37% Too High Choose this one
This system is optimal for buying the stock at the lowest price for the day.
We want to find the best position to stop automatically ignoring. Why 37%?
P(K) = Σ P(Being in position n) x P(Being chosen given in position n) 0 1 2 ... K K+1 K+2 ... N-1 N
We know the answer to
P(Being in position n) - it's
1/N as there are
N toilets, and we can select just 1.
Now, what is the chance we select them, given we're in position
The chance of selecting any of the toilets from
K is 0 - remember we're never going to buy then.
So let's move on to the toilets from K+1 and onwards. If K+1 is better than all before it, we have this:
P(Being in position n) = 1/N P(Being chosen given in position n) = 1
But, K+1 might not be the best price from all past and future prices. Maybe K+2 is better. Let's look at K+2
P(Being in position n) = 1/N P(Being chosen given in position n) = 1 - ( P(NOT being chosen given in position n) ) = 1 - ( 1/K+1 ) = K/K+1
K+2 we have
K+3 we have
So we have:
P(K) = 1/N x 0 + 1/N x 0 + ... + 1/N x 1 + 1/N x K/K+1 + 1/N x K/K+2 + ... + 1/N x K/N-1 = 0 + 0 + ... + (K/N) x (1/K + 1/K+1 + 1/K+2 ... + 1/N-1)
This is a close approximation of the area under 1/x - especially as x → ∞
N ∫ 1/x dx K N = [ln(x)] K = ln(N) - ln(K) = ln(N/K)
0 + 0 + ... + (K/N) x (1/K + 1/K+1 + 1/K+2 ... + 1/N-1) ≈ (K/N) x ln(N/K) and so
P(K) ≈ (K/N) x ln(N/K)
Now to simplify, say that
x = K/N
P(x) ≈ (x) x ln(1/x) ≈ -xln(x)
We can graph this, and find the maximum point so we know the maximum P(K) - or we can use calculus. Here's the graph:
Here's the calculus:
P'(K) = -ln(x) - x(1/x) = -ln(x) - 1 0 = -ln(x) - 1 -1 = ln(x) x = 1/e 1/e = 1/ 2.718281828 = 0.367879441 ≈ 0.37 = 37%
To apply this back to your situation with the stocks, if your stock updates every 30 seconds, and is open between 09:30 and 16:00, we have 6.5 hours = 390 minutes = 780 refreshes. You should keep track of the lowest price for the first 289 refreshes, and then buy your stock on the next best price.
x = K/N, the chance of you choosing the best price is 37%. However, the chance of you choosing better than the average stock is above 50% for the day. Remember, this method just tries to mean you don't loose money within the day - if you want to try to minimise losses within the whole trading period, you should scale this up, so you wait 37% of the trading period (e.g. 37% of 3 months) and then select.
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