2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 '14 at 9:28
  • $\begingroup$ 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? $\endgroup$ – Ginger Aug 20 '14 at 10:37
  • $\begingroup$ 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... $\endgroup$ – Ginger Aug 20 '14 at 10:42
  • $\begingroup$ or should Rt−Ry∼N(0,0.02*2)? $\endgroup$ – Ginger Aug 20 '14 at 10:46
  • 3
    $\begingroup$ I also did that sample question for a company beginning with G many moons ago ;)! $\endgroup$ – Chinny84 Aug 20 '14 at 20:44
6
$\begingroup$

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:

$$ P(R\geq 0.03) = 1 - P(R\leq 0.03) = 1 - N(0.03) = 1 - \Phi\left( \frac{0.03-\mu}{\sigma} \right) = 1 - \Phi(0.5) = 0.31 $$

where $\Phi$ is the standard normal distribution.

$\endgroup$
  • $\begingroup$ Aren't stockreturns R usually assumed to be log-normally distributed? $\endgroup$ – Meneldur Nov 27 '15 at 16:23
  • $\begingroup$ @Meneldur The log-returns are normally distributed and the stockprices are lognormally distributed. It follows from the assumed Geometric Brownian Motion where $S_t=S_0e^{rt+\sigma W_t}$. $\endgroup$ – emcor Nov 27 '15 at 17:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.