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In this paper (p. 36), authors wrote

$$p(n,T) = \Phi \Big(\frac{n}{T},\mu,\sigma \Big) - \Phi \Big (\frac{n-1}{T},\mu,\sigma \Big)\; (3) $$

Bellow we will use the approximation

$$p(n,T) = \frac{1}{T}N \Big(\frac{n}{T},\mu,\sigma \Big)\; (4)$$

On which $p(n,T)$ is their notation for $p(n = T)$, the probability that $n$ is $T$; $\Phi(\cdot)$ is the cdf and $N(\cdot)$ the pdf. I don't get why they use (4) to approximate (3), nor how they arrived to that equation.

Is this kind of approximation to the cdf standard? Sorry if the notation is a bit confusing. Let me know if I can improve the question to make it clearer.

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    $\begingroup$ Note that this is not actually an approximation to the CDF, but to the difference between two values of the CDF evaluated at different values of the argument. This is basically just a finite-difference approximation to the pdf, and if $T$ is "large", a reasonably good one at that. However, with the ease of finding software that will do the exact calculation for you, I don't know why the authors bother (although that may be made clear in the paper, which I didn't jump to.) $\endgroup$
    – jbowman
    Commented Apr 9, 2019 at 4:10
  • $\begingroup$ @jbowman Indeed, they use 4 to prove another thing later on, so yes, they bother for a reason. I'll try to figure out how they approximate pdf with the finite-difference. Thank you! $\endgroup$
    – jealcalat
    Commented Apr 9, 2019 at 4:36
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    $\begingroup$ See en.wikipedia.org/wiki/Mean_value_theorem. $\endgroup$
    – whuber ♦
    Commented Apr 9, 2019 at 13:30
  • $\begingroup$ @whuber Sorry, I don't follow you. How is my Q related with the m.v.t? $\endgroup$
    – jealcalat
    Commented Apr 9, 2019 at 19:09
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    $\begingroup$ Applied to your question, the Mean Value Theorem asserts there exists a number $0\le c\le 1$ such that $$p(n,T)=\frac{1}{T}\phi\left(\frac{n-c}{T},\mu,\sigma\right)$$ where $D_x\Phi(x,\mu,\sigma)=\phi(x,\mu,\sigma).$ When $\phi$ is continuous on the interval $[(n-1)/T,n/T],$ this implies you may use any such $c$ as an approximation and it gets better as $T$ grows larger (and $n/T$ stays about the same). $\endgroup$
    – whuber ♦
    Commented Apr 9, 2019 at 19:15

1 Answer 1

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This is a simple approximation where the integral over the density is approximated by taking the integrand to be equal to its upper value. Using a slight variation of their notation (to make the conditioning on parameters clearer), you have:

$$\begin{equation} \begin{aligned} p(n,T) &= \Phi \Big( \frac{n}{T} \Big| \mu, \sigma \Big) - \Phi \Big( \frac{n-1}{T} \Big| \mu, \sigma \Big) \\[6pt] &= \int \limits_{(n-1)/T}^{n/T} \text{N}(r | \mu, \sigma) \ dr \\[6pt] &\approx \int \limits_{(n-1)/T}^{n/T} \text{N} \Big( \frac{n}{T} \Big| \mu, \sigma \Big) \ dr \\[6pt] &= \text{N} \Big( \frac{n}{T} \Big| \mu, \sigma \Big) \int \limits_{(n-1)/T}^{n/T} \ dr \\[6pt] &= \Bigg[ \frac{n}{T} - \frac{n-1}{T} \Bigg] \cdot \text{N} \Big( \frac{n}{T} \Big| \mu, \sigma \Big). \\[6pt] &= \frac{1}{T} \cdot \text{N} \Big( \frac{n}{T} \Big| \mu, \sigma \Big). \\[6pt] \end{aligned} \end{equation}$$

Notice that the approximation step involves replacing the integrand density function with its value at the upper bound of the range of integration. So long as the range of integration is small (i.e., $T$ is large), the integrand will not vary much over the range of integration, so the approximation will be reasonably accurate.

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