# Why are these two methods to calculate standard deviation gives very different answers? [closed]

Values=[100, 101, 102.01, 103.03]

Method 1: Sum of the squared differences from the mean

Mean = 101.51

std = sqrt(((100 - 101.51)^2 + (101 - 101.51)^2 + (102 - 101.51)^2 + (103 - 101.51)^2) / 4) = 1.13

Method 2: Average of the natural log returns

returns = [1.01, 1.01, 1.01]

log_returns = [0.00995, 0.00995, 0.00995]

log_returns_squared = [0.000099009, 0.000099009, 0.000099009]

average = 0.000099009

sqrt average = 0.00995 which is 1 percent

One method gives 1.13 std and the other giver 1.

I am probably confusing two different things so could you please help me clarify this?

## 1 Answer

As far as I understand, you're calculating the standard deviation on two different things (prices and log-returns). Assume that the values (eg. stock prices) are defined by $$X_t$$, for $$t=1,\ldots,T$$. Then the first method described above, can be formulated as:

$$\begin{equation} \bar{\sigma} = \sqrt{\frac{1}{T}\sum_{i=t}^T (X_t - \bar{X})^2}, \end{equation}$$ which is the standard deviation of the stock prices, $$X_t$$, and is different from your second formulation. To see that, let $$r_t = \ln(X_{t}) - \ln(X_{t-1})$$ be your log-return at time $$t$$, then the second method can be described as: $$\begin{equation} \tilde{\sigma} = \sqrt{\frac{1}{T}\sum_{t=1}^T r_t^2}, \end{equation}$$ and calculates the standard deviation of the log-returns, $$r_t$$. The process $$(r_t)_{t \geq 0}$$ is recovered from differencing the log-transformation (log-prices) of the price process $$(X_t)_{t\geq0}$$ and therefore they are fundamentally different processes, hence giving you different standard deviations.

• Note that for the Black Scholes formula, you need the second definition i.e. $\tilde{\sigma}$. The first definition has some uses in finance, but relatively few. As Pleb said, they are very different things. – noob2 Feb 14 at 20:31