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?

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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 stockreturns $R$ are assumed normally distributed.

If market goes up 1%, the expected stockreturn is $R=\beta\cdot0.01=0.02$ (since $\beta$ being the senstivity to market).

Stockprice from $100$ over $103$ requires at least $103/100-1=0.03$ return $R$.

As we have now from the question $\sigma=0.02$ and $\mu=0.02$, with $R\sim N(\mu,\sigma)$ we get:

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

where $\Phi$ being the Standardnormal distribution.

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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