I asked the following question in CrossValidated, but I'm wondering if it is better suited to this platform...
If there are daily unconditional probabilities of an event that result in a weekly occurrence of an event of 25%, like so:
Day Prob
1 0.04026439
2 0.04026439
3 0.04026439
4 0.04026439
5 0.04026439
6 0.04026439
7 0.04026439
And I want to change these probabilities so they result in a weekly occurrence of 30% instead, I can use the formula:
$ P = 1 - (1 - rate)^{1/7} $
where rate is now 0.3.
However, what if the probabilities are not equal on all the days - for example, if there is a higher probability of the event occurring earlier in the week, like this:
Day Prob
1 0.0496772
2 0.04584501
3 0.04210299
4 0.04026439
5 0.03844668
6 0.03487194
7 0.03137491
The overall occurrence here is still 25%. But what if I want to modify these baseline probabilities so to reflect a given change in the weekly occurrence. What I want to do is to increase the daily baseline probabilities proportionally, so the shape of the baseline risk profile doesn't change but the overall occurrence does change - say from 25% to 30%.
Is it even possible?
Thanks
When you say "increase the daily baseline probabilities equally," in what sense do you mean "equally?" I presume that what you actually mean is proportionally, i.e., if $p_i$ is the event probability on day $i$ of the week, $i \in \{1, 2, \ldots, 7\}$, then you want to determine a constant $k$ such that $k p_i$ represents the updated probability on day $i$. In such a case, then what you can see is that the weekly event probability is $$1 - \prod_{i=1}^7 (1 - p_i) = 0.25,$$ so we require the updated weekly event probability to satisfy $$1 - \prod_{i=1}^7 (1 - kp_i) = 0.30.$$ This leads to a degree-$7$ polynomial in $k$, the approximate unique real solution of which is around $k \approx 1.23$. You would need a numeric method to solve this polynomial for general $p_i$; one approach is to employ Newton's method with initial guess $k_0 = 1$, or a more crude approach is a recursive bisection algorithm (i.e. secant method) with initial guesses $k_0 = 1$, $k_1 = 1/(\max_i p_i)$ in the case where the updated overall probability is strictly greater than the old overall probability, since we cannot have $k p_i > 1$. In the case where the updated probability needs to be lower, then of course $k_1 = 0$ would be your initial guess for the second endpoint in your recursive bisection.
Newton's method would be applied as follows. Suppose the desired overall probability is $p_T$. Construct the polynomial $$f(k) = (1-p_T) - \prod_{i=1}^7 (1 - kp_i),$$ and compute its derivative with respect to $k$: $$f'(k) = -\frac{d}{dk} \prod_{i=1}^7 (1 - kp_i) = - \prod_{i=1}^7 (1-kp_i) \sum_{k=1}^7 \frac{p_i}{1-kp_i}.$$ Then set up the recursion relation $$k_n = k_{n-1} - \frac{f(k_{n-1})}{f'(k_{n-1})},$$ with the initial guess $k_0 = 1$. Compute successive iterates $k_1, k_2, \ldots$, the limit of which should converge to the unique real root, given that none of the $p_i$ is equal to $1$.
Recursive bisection would work as follows. As in Newton's method above, construct $f$. Then for two initial guesses $k_0, k_1$, compute the midpoint $k_2 = (k_0 + k_1)/2$ and the values $$\left\{f(k_0), f(k_1), f(k_2)\right\}.$$ Compare the sign of $f(k_2)$ with the previous two values $f(k_0), f(k_1)$: for instance, if $f(k_2)$ has the opposite sign of $f(k_1)$, then the next guess $k_3$ is the midpoint of $k_1$ and $k_2$. Formally, $$k_{n+1} = \begin{cases} \frac{1}{2}(k_n + k_{n-1}), & s(f(k_n)) \ne s(f(k_{n-1})) \\ \frac{1}{2}(k_n + k_{n-2}), & s(f(k_n)) \ne s(f(k_{n-2})) \end{cases}$$ where $$s(x) = \begin{cases}1, & x > 0 \\ 0, & x = 0 \\ -1, & x < 0 \end{cases}$$ is the sign function. The idea is to always choose the next guess for $k$ in such a way that guarantees the root is located somewhere between any two successive iterates. Convergence is slower than with Newton's method but does not require computing derivatives.