Why does Bloomberg's HRH test the simple returns for normality?

Question

On a Bloomberg terminal, it is possible to use the HRH (Historical Return Histogram) function on individual assets. It basically generates a histogram of the (simple) returns and overlays them with a theoretical normal distribution, indicating whether the distribution of the (daily, weekly, monthly,...) returns is approximately normally distributed.

With a geometric Brownian motion model, we would assume that the log return is normally distributed and the (simple) return is lognormally distributed. Hence my question: Why does it test for normality and not lognormality?

Answer 1

For small changes, the log-return $\ln \frac{S_{t_i}}{S_{t_{i-1}}}$ is close to the simple return $\frac{S_{t_i}-S_{t_{i-1}}}{S_{t_{i-1}}}$: \begin{align*} \ln \frac{S_{t_i}}{S_{t_{i-1}}} &= \ln \Big(1+ \frac{S_{t_i}-S_{t_{i-1}}} {S_{t_{i-1}}} \Big)\\ &\approx \frac{S_{t_i}-S_{t_{i-1}}}{S_{t_{i-1}}}. \end{align*}

Note also that, assuming the SDE \begin{align*} \frac{dS_t}{S_t} = \mu dt + \sigma\, dW_t, \end{align*} then \begin{align*} \frac{S_{t_i}-S_{t_{i-1}}}{S_{t_{i-1}}} \approx \mu \Delta t_i + \sigma \Delta W_{t_i}, \end{align*} and \begin{align*} \ln \frac{S_{t_i}}{S_{t_{i-1}}} = \big(\mu -\frac{1}{2}\sigma^2\big) \Delta t_i + \sigma \Delta W_{t_i}, \end{align*} where $\Delta t_i=t_i-t_{i-1}$, and $\Delta W_{t_i} = W_{t_i}-W_{t_{i-1}}$ is normal.

That is, if the stock price is log-normally distributed, then the log-return is normally distributed, while the simple return is approximately normally distributed.