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.

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    $\begingroup$ Hi: You took the monthly sd and multiplied it by square root of 12. This was fine. But then you divided by the square root of 12 as if you still had 12 observations. As far as I can see, you have one observation ( because you grouped them all into one monthly number ) so you shouldn't be dividing by the square root of 12 also in the denominator. $\endgroup$
    – mark leeds
    Commented Jan 20, 2021 at 17:39
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    $\begingroup$ The number of observations are $T$. That should still be unchanged. $\endgroup$
    – phdstudent
    Commented Jan 20, 2021 at 17:40
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    $\begingroup$ The t test is invariant under linear transformation of the Data.Clearly,you are not transforming raw data but estimators -thereby introducing your own assumptions, no? $\endgroup$ Commented Jan 20, 2021 at 20:37
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    $\begingroup$ Hi: The number of observations is not T once you've made the decision to transform from monthly to yearly. You have 12 monthly observations but then you average them, thereby generating a monthly estimate which is ONE observation from a yearly standpoint. You can't have it both ways: claim that you have 12 yearly observations which is what you're doing with $\frac{\sigma_{r}^{annual}}{\sqrt(T)}$. $\endgroup$
    – mark leeds
    Commented Jan 20, 2021 at 22:21
  • $\begingroup$ Convert the monthly returns to annualized terms by multiplying by 12 before computing the mean and standard devation: your t-stat will not change. $\endgroup$ Commented Jan 21, 2021 at 0:22

1 Answer 1


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 $\sigma^2_{r_y} = T \sigma^2_{r}$.


$\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 $\sigma_{r_{y}}$).


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