I was unsure about the precise definition of "stock specific volatility". Used in this question "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?"

  • this questions appears in an existing thread (linked below) but I was unable to comment as I'm a new user and my answer post was removed because it was a question!

Probability of stock closing over a certain price

In the answer given in the thread above they assume that $\sigma^2$ = the conditional variance of the stock return given the benchmark

I would have thought that this would have been a more natural definition $\sigma^2$ = the marginal variance of the stock return

But if you then treat the stock and benchmark returns as a 2-variate normal and try to work out the conditional distribution for the stock given the benchmark then you don't have enough information. Need something like the benchmarks volatility as well. Can anyone clarify this for me please?

  • $\begingroup$ Thanks, that agrees with my first definition and makes the problem possible to answer which is a plus! $\endgroup$ Commented Jul 16, 2019 at 9:38

1 Answer 1


The stock specific volatility (also known as idiosyncratic volatility) is the volatility that remains after controlling for beta. I suppose you have $$R_i = R_f + \beta_i \cdot \big(R_m-R_f\big) + \varepsilon_i.$$ Then, the standard deviation of epsilon is your stock specific volatility. One frequently assumes $\varepsilon_i\sim N(0,\sigma^2_{\varepsilon_i})$. Then, the returns $R_i$ are normally distributed and $\mathbb{E}[R_i]=R_f + \beta_i \cdot \big(\mathbb{E}[R_m]-R_f\big)$. The variance is given by $\mathbb{V}\text{ar}[R_i]=\beta^2_i \cdot \sigma_m^2+\sigma_{\varepsilon_i}^2$.


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