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Suppose I have a series of gross P&L values, which are normally distributed with mean $\mu$, variance $\sigma^2$.

For positive P&L values, there is a $x\%$ commission. For example, $x=5\%$.

So the net pnl, $p_i$ is:

  • $p_i$ if $p_i$<0
  • $p_i * (1-x)$ if $p_i$>0

Given different $\mu%$ and $\sigma$, how can I calculate whether the net P&L over $n$ trades will be positive?

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  • $\begingroup$ I assume you are looking for the expected value of a trade, so it does not really make sense to look at "P&L over $n$ trades", right? $\endgroup$
    – MrLCh
    Commented May 14 at 16:41
  • $\begingroup$ OK. If so, what should I be doing? $\endgroup$ Commented May 15 at 10:57
  • $\begingroup$ A higher standard deviation will decrease the expected value. But I can't work out by how much. Intuitively, it might be something like (mu/signa) * (1-x) $\endgroup$ Commented May 15 at 12:19

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I believe there is no "simple" answer to your question. To calculate the expected value of net return $n$ you could do the following:

$$\begin{align}\mathbb{E}(n) &= \int_{-\infty}^{\infty} t * f_n(t) dt \\ &= \int_{-\infty}^{0} t * f_p(t) dt + \int_{0}^{\infty} t * (1-x) * f_p(t) dt \\ &= \int_{-\infty}^{0} t * f_p(t) dt + \int_{0}^{\infty} (t- tx) * f_p(t) dt \\ &= \int_{-\infty}^{0} t * f_p(t) dt + \int_{0}^{\infty} t * f_p(t) dt - x \int_{0}^{\infty} t * f_p(t) dt \\ &= \mathbb{E}(p) - x \int_{0}^{\infty} t * f_p(t) dt \end{align}$$

where $f_n$ denotes the density of the $n$ and $f_p$ denotes the density of $p$ (so the density of the normal distribution with mean $\mu$ and variance $\sigma^2$).

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    $\begingroup$ If you expand this out you can clearly see the expectation is negative (if mu is zero), which makes sense because you are paying away comission on a RV which has expectation zero. Further you can observe that the expectation is exactly mu - 0.5x $\endgroup$
    – Attack68
    Commented May 15 at 18:46
  • $\begingroup$ You are right with the expansion. Just edited my answer to recover the expected value of p, but I do not see where you get the -0.5x. $\endgroup$
    – MrLCh
    Commented May 15 at 19:07
  • $\begingroup$ (yes i was using symmetry with mu =0. when it is not zero doesnt look like a closed form) $\endgroup$
    – Attack68
    Commented May 15 at 19:19

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