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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?

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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.

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    $\begingroup$ 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. $\endgroup$ – noob2 Feb 14 at 20:31

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