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• Hey Rune, thanks for your great work! Have a question about log returns and the „additive advantage“. I don‘t quite get it: Couldn‘t I use returns and cumprod() instead of log returns and cumsum(), and get the same effect? Thx!

  • Hey Volker.
    Sure it is the same. The difference is that sum is faster than product.
    If the difference in these problems is big I didn’t try. But it is custom to use sum for the above reason.

• Dear Rune. Thanks you for this excelent work. I have a doubt with this sentence

  data['Position'] = (data['Signal'].apply(np.sign) + 1)/2

  I undestand that the fucntion np.sign(), return the next:

  np.sign(x) -1 if signal >= 0, for so much "data['Position']" will be (-1+1)/2 = 0
  np.sign(x) 0 if signal == 0 , for so much "data['Position']" will be (0+1)/2 = 0.5 ¿?
  np.sign(x) 1 if signal <= 0, for so much "data['Position']"will be (1+1)/2 = 1

  If we are looking for data['Position'] 0 y 1, please explain this sentence to us. Cordial greetings

  • Let's try to break it down.

    
data['Signal'] = data['Adj Close'].rolling(5).mean() - data['Adj Close'].rolling(20).mean()


    In signal we have the rolling average of 5 days minus rolling average of 20 days.

    We want to use that as a signal.

    data['Position'] = (data['Signal'].apply(np.sign) + 1)/2

    Let's break that down too.

    data['Signal'].apply(np.sign)

    This gives the sign of the signal: -1 if negative, 1 if positive (0 if 0).

    + 1

    Says, if -1 we get 0 and 2 if 1.

    Then /2

    Ensure we have 0 if short and 1 if long.

    That is what we try to do in step 2.

    This results in a Signal line, which is the differences of the two moving averages. When the signal line is positive our position is 1 (buy and hold) otherwise 0 (sell).

• Dear Rune
  I keep trying to understand these sentences. In this case, I do not understand :

  data[['Log return', 'Return']].cumsum().apply(np.exp)
  np.exp(data[['Log return', 'Return']].mean()*252)

  ¿The last row of the DataFrame data[['Log return', 'Return']].cumsum()

  Isn't the final result of the purchase/sales strategy and the strategy of maintaining the position?. ¿For what is necessary to apply the np.exp() function.

  I need to understand it to be able to trust the result. Thanks in advance. cordial greetings

  • Let's try to break it down.

    data['Return'] = data['Position'].shift(1)*data['Log return']

    Keeps the return of our investment. It keeps the log-return for each day we have invested. The shift is because we need one day to react. We decide after closing hours to hold, buy, or sell. If we hold - we have a 1 in our position. Then we get the log return. otherwise, we have 0 and get nothing (0).

    data[[‘Log return’, ‘Return’]].cumsum().apply(np.exp)
    This then shows what we get if we had the investment all the time (log return) vs our investment strategy (return).
    The cumsum() is summing up all log-returns (in both log-return and return) and the we apply np.exp to get the actual return. This is the beauty of calculating with log returns.

    The mean()*252 is just used to get an annualized return.

    Please feel free to ask any clarifying questions.

    /Rune

• Thanks Rune

  In yours book "Python for Financial Analisys", this question is is perfectly explained.

  In the chapter "Log returns " we can see 𝑎×𝑏=exp(𝑙𝑜𝑔(𝑎×𝑏))=exp(log(𝑎)+log(𝑏))

  • Awesome - thanks for sharing

• Dear Rune
  I keep learning with your script. I have another question that I will thank you for clarifying.

  In the statement "data [['Log return', 'Return']]. Cumsum (). Apply (np.exp)", you apply the function np.exp () to the two columns.

  I understand that we may use: "cumsum" with "return" and "cumprod" with "log return", and that "np.exp()", It is only applicable to logarithms .

  Can you confirm that your sentence is correct and what is my mistake? I appreciate your help in advance.

  • I think you are right.
    Let us see.
    The log returns are used to make it possible to add returns together.
    The np.exp is to transform log-returns to %-returns.
    Hence the cumsum().apply(np.exp) is adding log-returns and transforms them to %-return.
    If you had %-returns it would be:
    cumprod() directly. But this is slower and the standard in finance is to use log-returns in these cases.
    Hope it helps,
    Rune

• Dear Rune:

  Computers, Artificial Intelligence, Machine Learning with its algorithms, geopolitics, etc., have so complicated the financial system, which sometimes can't understand today.

  For this reason, to the people who cannot trust in a long way for their investments, the classic strategies do not serve. In my opinion, it is necessary to look for stock portfolio management strategies that allow us to preserve it in the worst circumstance that exists and achieve a benefit that even surpasses the cost of life index.

  As I learn this illusive language, I am committed to searching for some strategy with this objective. I would like to count with your authorization to use, adapting them, some sentences of this script.

  cordial greetings

  • I believe you are right. It is very difficult to make a bullet proof strategy. In this tutorial we only focus on backtesting in a classical way.
    I am interested to add ML to the strategy in new ways and test it.

• Rune, I've been stubborn, delete the last query, please. Thanks

  • It is done. Sorry I was a bit slow to reply.

• Dear Rune.
  In the statement 'data['Position'] = (data['Signal'].apply(np.sign) + 1)/2', I see the following problem. When 'signal' is equal to 0, this statement will return 0.5 in 'position'.
  To make these 0.5 values 0, I have done the following.

  # Order number of records with 0.5 in position
  data.reset_index(inplace=True)
  lista = np.where(data['position'] == 0.5)
  # Change 0.5 to 0
  for each_value in lista:
  data.loc[each_value, 'position'] = 0

  What do you think about this? Am I wrong?

  • Hi Efueyo,

    You might be right in the special case of signal == 0.

    Thank you for pointing it out.

  • This is one solution.
    # # Pass the value from 0.5 to 0
    df.loc[data['Position'] == 0.5, 'Position'] = 0