A stock has beta of 2.0 and stock specific daily volatility of 0.02. Suppose that yesterday's closing price was 100 and today the market goes up by 1%. What's the probability of today's closing price being at least 103?
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$\begingroup$ Hi Ginger, welcome to quant.SE! I've removed your 'disclaimer' and cleared up the title. However, I believe one thing is missing: what model are you using? $\endgroup$– Bob Jansen ♦Aug 20, 2014 at 9:28
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$\begingroup$ Hi, Rob, thanks! What model should be using here, this is the question I am thinking of. This is an interview question. Since I am new so I thought there is a classical model for this problem, is it? $\endgroup$– GingerAug 20, 2014 at 10:37
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$\begingroup$ If I can choose the model, I would do it like this, R_t-R_y is normal distribution, R_t is today's closing price, R_y=100*(1+1%)=101. and $R_t-R_y\sim N(0,0.02)$. Then beta of 2.0 would be useless... $\endgroup$– GingerAug 20, 2014 at 10:42
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$\begingroup$ or should Rt−Ry∼N(0,0.02*2)? $\endgroup$– GingerAug 20, 2014 at 10:46
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4$\begingroup$ I also did that sample question for a company beginning with G many moons ago ;)! $\endgroup$– Chinny84Aug 20, 2014 at 20:44
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1 Answer
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Usually stock returns are assumed to be normally distributed: $R\sim N(\mu,\sigma)$
If market goes up 1%, the expected stock return is $\mu = \beta\cdot 0.01 = 0.02$ (since β is the senstivity to market).
Stock price going from 100 to over 103 requires a return $R$ of at least 103/100 – 1 = 0.03.
As we have from the question σ = 0.02, we get:
$$ P(R\geq 0.03) = 1 - P(R\leq 0.03) = 1 - F(0.03) = 1 - \Phi\left( \frac{0.03-\mu}{\sigma} \right) = 1 - \Phi(0.5) = 0.31 $$
where $F$ is the generic normal cumulative distribution function and $\Phi$ is the cumulative distribution function for the standard normal distribution.
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$\begingroup$ Aren't stockreturns R usually assumed to be log-normally distributed? $\endgroup$– MeneldurNov 27, 2015 at 16:23
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1$\begingroup$ @Meneldur The log-returns are normally distributed and the stockprices are lognormally distributed. It follows from the assumed Geometric Brownian Motion where $S_t=S_0e^{rt+\sigma W_t}$. $\endgroup$– emcorNov 27, 2015 at 17:15
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$\begingroup$ Hello, sorry for waking up the question after years .. why is the stock expected return equal to $\beta\cdot 0.01$? $\beta$ links the stock expected return with the market expected return, and 0.01 is today's market return, not the market expected return. $\endgroup$– marcoNov 14, 2020 at 20:19