# Distribution when positive values are rescaled?

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?

• 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? May 14 at 16:41
• OK. If so, what should I be doing? May 15 at 10:57
• 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) May 15 at 12:19

## 1 Answer

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 $$\sigma^2$$).

• 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
– Attack68
May 15 at 18:46
• 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. May 15 at 19:07
• (yes i was using symmetry with mu =0. when it is not zero doesnt look like a closed form)
– Attack68
May 15 at 19:19