Annualizing Monthly Returns Without Distorting a Mean Test

# T-statistics on monthly returns vs annualized monthly returns


# T-statistics on monthly returns vs annualized monthly returns












eqI am very confused about a very basic question. This is probably more statistics than quantitative finance, but still, should be useful for this stackexchange board as well.

Let's assume I have monthly returns $r_1, r_2, ...., r_T$.

I can compute average return and standard deviation as:

\begin{equation} \bar{r} = \frac{1}{T} \sum_{t=1}^T r_i \end{equation}

I can compute standard deviation of return as: \begin{equation} \sigma_r = \sqrt{\frac{1}{T-1} \sum_{t=1}^T(r_i - \bar{r})^2} \end{equation}

Thus if I want to test whether the mean is equal to zero I can build the t-statistic:

\begin{equation} t-stat = \frac{\bar{r}}{\sigma_r/\sqrt{T}} \end{equation}

Now assume I annualize the mean and the standard deviation by making:

\begin{equation} \bar{r}^{annual} = \bar{r} \times 12 \end{equation}

\begin{equation} \sigma_{r}^{annual} = \sigma_r \times \sqrt{12} \end{equation}

Now if I compute the same t-stat, as above I get:

\begin{equation} t-stat^{annual} = \frac{\bar{r}^{annual}}{\sigma^{annual}_r/\sqrt{T}} = \sqrt{12} (t-stat) \end{equation}

My question is what am I doing wroing? Why is the t-stat becoming multiplied by sqrt(12)? If instead I first annualize the returns series $r_i$ by making $r_i \times 12$ this doesn't happen, and I get the same t-stat, regardless of whether I annualize or not.

## Answer by mark leeds (score 1)

https://quant.stackexchange.com/a/60613

This is a hopefully clearer explanation of what I've been saying in my comments. The background is that you have an average monthly return $\bar{r}_{m} = \frac{1}{12}\sum_{i=1}^{12} r_{i}$ and a monthly variance estimate $\sigma^2_{\bar{r}_{m}}$ where $\sigma^2_{\bar{r}_{m}} = \frac{\sigma^2_{r}}{12}$. In what follows, it is assumed that these estimates have already been computed.

Now, we want to convert from monthly to yearly ( we assume that the returns can be added because they are small enough ) to get the yearly return and yearly variance, so we need to make the assumption that the monthly returns are iid.

So, we want to compute

$r_{y} = \sum_{i=1}^{T} \bar{r}_m$ where $T = 12$ and $\sigma^2_{r_y} = \sum_{i=1}^T \sigma^2_{r}$ where $T = 12$.

So, after using the iid assumption and some algebra, it ends up being straightforward to show that $r_y = T \bar{r}_{m}$ and $\sigma^2_{r_y} = T \sigma^2_{r}$.

Therefore,

$\frac{r_y}{\sigma_{r_y}} = $ $ \frac{T \bar{r}_{m}}{\sqrt{T} \sigma_{r}} $ $ = \frac{\bar{r}_{m}}{\frac{\sigma_{r}}{\sqrt{T}}}$

The last term is the same estimator that the OP has defined as $t - tstat$. Finally, note that this statistic would be compared to the t-distribution with $T-1$ degrees of freedom because $\sigma_r$ consists of $T$ observations (not $\sigma_{r_{y}}$).