Realized Variance is written as $RV_{[0,T]}^{n} = \sum_{j = 1}^{n} r_{j,n}^2$, where $r_{j,n}$ is the log return for the $j$th increment, and $n$ is the total number of sample points in the time period $[0,T]$.

Integrated Variance (also Quadratic Variation in the case of a geometric Brownian motion stock price model), is $IV = \int_0^{T}\sigma^2 dt$, where $T$ is the total number of years.

I know that as we have more data points ($n \to \infty$), that $RV \to IV$ in probability. This makes sense to me as $IV$ can be written $\int_0^T(\text{d}\ln S)^2$, and so if we have an infinite number of sampling points in $[0,T]$ then $RV$ is the sum of the square of log returns over infinitesimal increments.

I have seen that with variance swaps (not sure if my understanding is correct) that realized variance can be approximated for $n$ days as $RV = \sum_{j = 1}^{n} r_{j}^2$. I can see this as a very crude approximation of $\int_0^T(\text{d}\ln S)^2$, where $T = \frac{n}{252}$, where there are very few sampling points, but am wondering what kind of approximation could be made to show that the $IV \approx RV$.

I am trying to see if I can get an intuitive understanding of how such a crude approximation could possibly be close to the actual $IV$, but am having trouble. For example, with a Riemann sum and a 'normal' integral from Calculus, $\int_0^L f(x)\,dx$ can be approximated using $\sum_{i=1}^{n}{f(x_i)\Delta x} = \sum_{i=1}^{n} \frac{L}{n}f(iL/n)$, and this makes sense to me as we are just assuming $f(x)$ is constant over the increments.

While trying to do something similar with $RV$ and $IV$, I break up $IV = \int_0^T(\text{d}\ln S)^2 = \sum_{i = 1}^n\int_{{\frac{i-1}{252}}}^{\frac{i}{252}}(\text{d}\ln S)^2$, for each day involved. Then I can directly compare with the $RV$ approximation $\sum_{j = 1}^{n} r_{j}^2$, and for this approximation to make sense I am trying to show that $\int_{{\frac{i-1}{252}}}^{\frac{i}{252}}(\text{d}\ln S)^2 \approx r_i^2$, under some sort of simplification. If this was an integral with the $\int f(x)dx$ or even $\int df(x)$, I can see how to simplify and show that they are approximately equal, but I don't know what to do with $\int_{{\frac{i-1}{252}}}^{\frac{i}{252}}(\text{d}\ln S)^2$, since the increment of the log stock price is squared.

I tried writing $r_i = \ln S_{\frac{i}{252}} - \ln S_{\frac{i-1}{252}}$ but would only be able to reach a straight forward simplification if $\text{d} \ln S$ was not squared. Here $r_i^2 = (\int_{{\frac{i-1}{252}}}^{\frac{i}{252}}(\text{d}\ln S))^2$ and this is clearly not the same as $\int_{{\frac{i-1}{252}}}^{\frac{i}{252}}(\text{d}\ln S)^2$. I am wondering if there is some valid approximation that would make this be true, but can't seem to justify it.

I'd appreciate some help in trying to reach a valid approximation. I know that as the mesh size becomes finer that the two should converge, but I wanted to get an idea of how we could intuitively show that $RV$ is an approximation of $IV$ similar to how a Riemann sum is comparable to an Integral. Thanks in advance!


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