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I am reading Paul Wilmott's Frequently Asked Questions in Quantitative Finance, and there is a question that states the following:

Every day a trader either makes 50% with prob- ability 0.6 or loses 50% with probability 0.4. What is the probability the trader will be ahead at the end of a year, 260 trading days? Over what number of days does the trader have the maximum probability of making money?

He explains how to break even the trader's best chance is at ~164 days. This is simple, he solves for $1.5^n 0.5^{260-n}$. But then he says that the trader's average return per day is:

$1−e^{0.6 \ln1.5 + 0.4\ln0.5}$ = −3.34%

Where does this formula come from? Any idea how to get the formula on the left hand side?

JoeVictor
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    The equation you have is backwards. He starts with $1$ and ends with the exponential, which is about $0.9666$. You should subtract $1$ from the exponential, which gives the claimed negative result. – Ross Millikan Jul 04 '20 at 20:14

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The $164$ days is the number of profitable days that are needed out of $260$ trading days for the trader to break even. It comes from solving $1.5^n\cdot 0.5^{260-n}=1$. You did not include the rest of the equation. We can say the expected number of up days is $0.6 \cdot 260=156$. This shows that his expectation is negative, as he needs more than the expected number of up days to break even.

To get the exponential, consider the log of the amount of the money the trader has. It starts at $0$ because he has $1$ unit. After one day, the expected value of the log is $0.6 \log 1.5 + 0.4 \log 0.5$. The expected value of his account after one day is $e^{0.6 \log 1.5 + 0.4 \log 0.5}$ so the expected gain is this minus $1$, which comes out $-0.0334$

Ross Millikan
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  • why are we dealing with the log and not just the direct expectation: $E[return] = 0.60.5 + 0.4(-0.5)$ ? – JoeVictor Jul 05 '20 at 23:07
  • @QuantumHoneybees: because the returns are multiplicative, not additive. If he wins one and loses one, his bankroll is $1\cdot 1.5 \cdot 0.5=0.75,$ not $1-0.50+0.50=1$ – Ross Millikan Jul 05 '20 at 23:52
  • That sort-of makes sense? But then why is it that in another book, the question is approached additively? – JoeVictor Jul 05 '20 at 23:54
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    The critical thing is the question you ask. Say there are only two trading days. If both are up, probability $0.36$ you have $2.25$. If one is up and one is down, probability $0.48$ you have $0.75$. If both are down, probability $0.16$ you have $0.25$. This questions asks the chance you are ahead, which here is $\frac 14$. The expected value is $1.21$. On an expected value basis, it is a good bet and you should take it. Over the long run, the expected value comes from rare events that you win a lot of money. – Ross Millikan Jul 06 '20 at 02:07
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    For the extreme version of this, see this It takes a strange utility function to buy insurance and play the lottery. – Ross Millikan Jul 06 '20 at 02:07
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I do not follow where the confusion is. Let's say he had 1 rupee at the beginning of the day. He has probability of 0.6 to end with 1.5 rupee (50% gain) and 0.4 probability to end with 0.5 rupee (50% loss).

So at the end of the day, his money on an average will be $= (1.5)^{0.6}.(0.5)^{0.4} \approx 0.9666.$

He started the day with 1 rupee so his average daily return $= 0.9666-1 = -0.0334 = -3.34 \% $

Math Lover
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  • your last formula is correct but the other way around. It's $0.09666 - 1$. But that made it click. Thanks! – JoeVictor Jul 04 '20 at 23:25
  • Sorry for the typo. I fixed it now. – Math Lover Jul 05 '20 at 04:39
  • Wait actually, I still don't get this. Shouldn't the expected return per day be 0.5 * 0.6 + (-0.5)*0.4? Why are we putting the probability in the exponent? What rule of probability theory is this? – JoeVictor Jul 05 '20 at 23:02
  • You are right about the expected return if you take one day in isolation. But if you are figuring out average daily return over a period of time, you will have to keep multiplying. For example, how do you calculate return on your investment over a long period of time? $P(t) = P(0)e^{rt}$ where P(t) is principal after time t and r is the annual interest rate. Please read https://en.wikipedia.org/wiki/Compound_interest#Continuous_compounding. Please also read Kelly Criterion in the context of why we are losing money in this case in long run. – Math Lover Jul 06 '20 at 05:32
  • One more thing, Over what number of days does the trader have the maximum probability of making money? Your answer of 164 days is not correct. That is the break even point in terms of how many days he should make money vs. lose to overall make money. But the probability of him making money on 164 days is pretty bad which means in long run he would lose money. Btw his chances of making money is highest after 3 days as per my calculation. – Math Lover Jul 06 '20 at 05:38
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$$e^{0.6\ln 1.5+0.4\ln 0.5}=e^{0.6\ln 1.5}\cdot e^{0.4\ln 0.5}=(1.5)^{0.6}\cdot (0.5)^{0.4}=0.966\ldots\quad(\because e^{n\ln x}=e^{\ln x^n}=x^n)$$ times the money at the beginning of the day becomes the money at the end of the day on average because the probability of getting $1.5$ times the money at the end of the day is $0.6$ and that of getting $0.5$ times the money is $0.4$. So, over a $10$(say)-day period, money gets $(1.5)^{6}\cdot (0.5)^{4}$ times (just a check).

So, the average loss per day is $[1-(1.5)^{0.6}\cdot (0.5)^{0.4}]\%$.

Also, it seems to me that $164$ are the days required in the best case to achieve a no-profit-no-loss situation in $260$ days. Hence, $1.5^{n}\cdot0.5^{260-n}=1\Rightarrow n=164$. Note that the best case is far from from the $10$-day logic as $(1.5)^{0.6\cdot260}(0.5)^{0.4\cdot260}\approx 0$.

Sameer Baheti
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  • This is taken from a book, so I don't understand it either. I can see that it's positive 3.34% when calculated out, but the book says that's it's negative. What is the RHS of this equation? I'm still lost and don't quite understand what you wrote – JoeVictor Jul 04 '20 at 20:01
  • @QuantumHoneybees What exactly you didn't follow? – Sameer Baheti Jul 04 '20 at 20:07
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    Did you do the computation? $1.5^{0.6} \cdot 0.5^{0.4}\approx 0.9666$. Yes, the probability of a profit is more, but the profits are smaller. – Ross Millikan Jul 04 '20 at 20:13
  • @Ross Millikan Thanks! I edited my answer. – Sameer Baheti Jul 04 '20 at 20:17