# Probability of stock closing over a certain price

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?

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

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.

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