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You are a trader considering a stock whose next price movement will be up or down with equal probability. Fortunately, you have access to a binary signal that is predictive of the price change (and observable ahead of time). The signal is 60% accurate: conditional on the stock going up, there's a 0.6 probability that the signal says up (and 0.4 probability that it says down); the reverse is true if the stock goes down.

You have access to many such signals, all with the same accuracy. Assume they are conditionally independent given the price movement (e.g. conditional on the stock going up, there's a 0.6^2 probability that signals one and two both point up). Suppose you observe N+1 signals pointing up and N pointing down (out of a total of 2N+1 signals, i.e. the total number must be odd). What's the probability that the stock will go up?

Himanshu Agarwal
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  • This is just a fairly straightforward exercise in the application of Bayes's theorem. You'll get more benefit from trying to do the calculation yourself first, and then seeking help if you get stuck, than from someone spoon-feeding you details of the entire procedure. As a check, my calculation gives me a probability of $\ 0.6\ $ that the stock price will go up. – lonza leggiera Mar 04 '19 at 10:52
  • Lonza, I understand Bayes's theorem. And solved various questions on the same. Just I am stuck on this one. Can you share your mathematical expression please – Himanshu Agarwal Mar 04 '19 at 11:25

1 Answers1

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Let $\ M\ $ be the random variable indicating the movement in the stock price, $\ S_i, i = 1, 2, \dots, 2\, N+1\ $, the signals, and $\ U, D\ $ the sets of indices $\ i\ $ for which the signal $\ S_i\ $ has the value $\ u\ $ (for "up") or $\ d\ $ (for "down"), respectively. We are told that $\ \vert U\vert = N+1\ $ and $\ \vert D\vert = N\ $.

\begin{eqnarray} \ & &\mathrm{P}\left(\,M = u\,\vert\, S_i = u\ \mathrm{for}\ i\in U\ \&\ S_i = d\ \mathrm{for}\ i\in D\, \right) \\ &=& \frac{\mathrm{P}\left(\,M = u\,\&\, S_i = u\ \mathrm{for}\ i\in U\ \&\ S_i = d\ \mathrm{for}\ i\in D\, \right)}{\mathrm{P}\left(\, S_i = u\ \mathrm{for}\ i\in U\ \&\ S_i = d\ \mathrm{for}\ i\in D\, \right)}\\ &=& \frac{\mathrm{P}\left(\, S_i = u\ \mathrm{for}\ i\in U\ \&\ S_i = d\ \mathrm{for}\ i\in D\,\vert\,M=u\, \right)\,\mathrm{P}\left(\,M=u\,\right)}{\mathrm{P}\left(\, S_i = u\ \mathrm{for}\ i\in U\ \&\ S_i = d\ \mathrm{for}\ i\in D\, \right)}\\ &=& \frac{0.6^{N+1}\,0.4^N\,0.5}{\mathrm{P}\left(\, S_i = u\ \mathrm{for}\ i\in U\ \&\ S_i = d\ \mathrm{for}\ i\in D\, \right)}\ . \end{eqnarray} For the denominator, we have \begin{eqnarray} & &\mathrm{P}\left(\, S_i = u\ \mathrm{for}\ i\in U\ \&\ S_i = d\ \mathrm{for}\ i\in D\, \right)\\ &=& \mathrm{P}\left(\, S_i = u\ \mathrm{for}\ i\in U\ \&\ S_i = d\ \mathrm{for}\ i\in D\,\vert\,M=u \right)\mathrm{P}\left(\,M=u\,\right)\\ & & +\, \mathrm{P}\left(\, S_i = u\ \mathrm{for}\ i\in U\ \&\ S_i = d\ \mathrm{for}\ i\in D\,\vert\,M=d \right)\mathrm{P}\left(\,M=d\,\right)\\ &=& 0.6^{N+1}\,0.4^N\,0.5 + 0.6^N\,0.4^{N+1}\,0.5\\ &=& 0.6^N\,0.4^N\,0.5\ . \end{eqnarray} So finally, \begin{eqnarray} \ & &\mathrm{P}\left(\,M = u\,\vert\, S_i = u\ \mathrm{for}\ i\in U\ \&\ S_i = d\ \mathrm{for}\ i\in D\, \right)\\ &=& \frac{0.6^{N+1}\,0.4^N\,0.5}{0.6^N\,0.4^N\,0.5} = 0.6 \end{eqnarray}

lonza leggiera
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