# Add transaction costs to prediction

An algorithm predicts price movement by some certainty and it invests proportional to the confidence level. Predictions range from -1 to +1, -1 meaning sell for a value of $1 +1 meaning buy for a value of$1. Then the profit is calculated by multiplying the prediction with the relative price movement of the security traded.

Now assume a transaction cost of 0.6%. How does that change the profit the algorithm makes? For now, we only calculate the transaction cost for one cycle, i.e buy or sell once and the next time step sell or buy again in order to realize the profit.

So to clarify. I have two variables pred which is a prediction ranging from -1 and +1. I also have d_price which is the relative price movement of the security. This can be 0.0003 or -0.002 or something similar. You calculate this by d_price = (price_t1 - price_t0) / price_t0

I have this eqution now:

profit = pred * d_price

The algorithm makes two trades. It makes a trade when it makes the prediction at time step t0, then it makes another trade at time step t1 in order to realize a profit. So if it predicts +0.5 and then the relative price movement is +0.01 then the profit it makes is 0.005.

What I'm asking about is how this changes when there is a transaction cost of trans=0.006. The transaction cost if percentage based, meaning if I buy 1 amount, I will receive 0.994 only. Likewise, if I sell 1 amount I will receive price * 0.994

profit = f(pred,d_price,trans)

What is f ?

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What exactly is your question? Obviously, positive transactions costs will reduce your profit. –  Louis Marascio May 16 '13 at 12:07
@LouisMarascio the question is by how much –  siamii May 16 '13 at 13:08
@siamii, you are asking a question that is impossible to answer without a host of other information which you did not provide. –  Matt Wolf May 16 '13 at 13:17
@Freddy I've made some amendments. See if it is clearer now? –  siamii May 16 '13 at 16:20