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Equation 6 (right truncation) #51

Open athowes opened 2 months ago

athowes commented 2 months ago

Hi! Reading through this at the moment and feel like I'm getting something different to what you have for Equation (6) and (7). Here would be my version, where I think I'm just applying $P(A, B) = P(A | B) P(B)$:

$$ \begin{align} \mathbb{P}(S = P + \tau | P = p, S < T) &= \frac{\mathbb{P}(S = P + \tau, P = p, S < T)}{\mathbb{P}(P = p, S < T)} \ &= \frac{\mathbb{P}(S = P + \tau < T | P = p)\mathbb{P}(P = p)}{\mathbb{P}(S < T | P = p)\mathbb{P}(P = p)} \ &= \frac{\mathbb{P}(S = P + \tau < T | P = p)}{\mathbb{P}(S < T | P = p)} \ &= \cdots \end{align} $$

I'm not sure, but I might guess that what you have written isn't right as the denominator of Equation (7) contains $\tau$ but then in Equation (8) and (9) there is no dependence on $\tau$? The version above does not have dependence on $\tau$ in the denominator, compatible with $\int_0^{T-p} f_p(x) \text{d}x$ being an integral over all possible positive delays less than $T - p$. Should Equation (9) have some constraints on it too? Like $p + \tau < T$ else expression is zero?

Let me know if I'm missing something!

Edit: also for Equation (18) don't you also need $S < T$ in the numerator?