Supposoe we have normal variable $ N(\mu_t, \sigma_t^2 ) $.
Let's define X = $\int_0^T N(\mu_t, \sigma_t^2 ) dt$.
Is X also a normal variable?
It's simple to check that expectation of X is simply $\int_0^T \mu_t dt$ as $\mathbb{E} (\int) = \int (\mathbb{E})$.
BUT it is not the case for variance as
$\mathbb{V} (X) = \mathbb{V} (\int_0^T N(\mu_t, \sigma_t^2 ) dt) \approx \mathbb{V} (\sum_i N(\mu_i, \sigma_i^2 ) \triangle) = > \mathbb{V} (\sum_i N(\mu_i, \triangle^2\sigma_i^2 ) )$
Entering the sum into normal variable:
$\mathbb{V} (N(\ \sum_i \mu_i, \ \sum_i \triangle^2\sigma_i^2 \ ) ) = \sum_i \triangle^2\sigma_i^2$
Now I would like to get back to integrals and do something like this:
$\sum_i \triangle^2\sigma_i^2 \approx \int_0^T \triangle \sigma_t^2 dt, \ \triangle \to 0$
But it seems weird...
Could someone help me to find the good variance and figure out where I was wrong?