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I have been investigating the following recurrence that pops out in describing the probability a particular Monte Carlo algorithm for min cuts succeeds:

$$P_k = P_{k-1} - P_{k-1}^2 / 4$$

where $P_0 = c$ for some $0 < c < 1$. In my case, all I need is a lower bound for $P_k$ but ultimately I want it to be asymptotically as tight as I can get because this success probability plays a role in how many times I will re-run the algorithm to make it much less likely the algorithm fails. A naive approach uses the fact that $0 \leq P_{k-1} \leq 1$ to make the recurrence become

$$P_{k} \geq P_{k-1}\left(1 - 1/4\right) = \frac{3}{4} P_{k-1} \geq \left(3/4\right)^{k} P_{0}$$

The depth of the recursion tree in the algorithm is $d = c \log(n)$ for some constant $c$ which makes the above bound state that $P_{d} = \Omega(1/n^b)$ for some constant $b$. This is not great because it can be shown that $P_{d} = \Omega(1/\log(n))$. From the original paper on the min cut algorithm, they tackled this recurrence by making $P_k = 4/Z_k$ for some new variable $Z_k$. With this, one can then find that $P_k = \Omega(1/k)$ and so the desired result falls out.

I asked a question recently here to try and get some intuition into this transformation because it was not intuitive to me. One of the simpler approaches mentioned in the answer effectively treated the recurrence as a finite difference approximation to some differential equation and used that idea to spit out a differential equation we could work on instead. By this approach, one can find that the related diffeq is $\dot{p} = -p/4$ with $p(0) = c$ and it gives us that $p(k) = 4/(4c^{-1} + k)$. Interestingly, this $p(k)$ is asymptotically correct when you compare to the $P_k = \Omega(1/k)$ result.

My question is, is there any rigorous way to relate the answer to a recurrence of the form $x_{k} = x_{k-1} + f(x_{k-1})$ for some function $f$ to the solution you get from the differential equation $\dot{x}(t) = f(x(t))$? In particular, are there any well known results that give inequalities that relate the two so I can find lower bounds or upper bounds of a recurrence using the corresponding differential equation (and vice versa)?

Edit

More specifically, if we have the recurrence $X_{k} = X_{k-1} + f(X_{k-1})$ with $X_{0} = c$ and differential equation $\dot{x}(t) = f(x(t))$ with $x(0) = c$, do we have any bound of the form $K_1 x(k) + d_1 \leq X_{k} \leq K_2 x(k) + d_2$ for all $k \geq 0$ with constants $K_1, d_1, K_2, d_2$?

spektr
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2 Answers2

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The calculus argument is that you have $\frac{dx}{dk}\approx -\frac{x^2}{4}$ suggesting $\int -\frac1{x^2}\, dx \approx \int \frac14\, dk$ i.e. $\frac{1}{x}\approx \frac{k}{4}+B$ for some constant $B$, i.e. $x\approx\frac{1}{\frac{k}{4}+B}$. But here you have a discrete recursion so may need some adjustment.

When $0 < P_0=c \le 1$ you have the bounds $$\dfrac{1}{\frac k{3} + \frac1c} \le P_k \le \dfrac{1}{\frac k4 + \frac1c}$$ or if you dislike fractions of fractions, $$\dfrac{3c}{kc + 3} \le P_k \le \dfrac{4c}{kc + 4}$$ and this can be shown by induction, since $x-\frac{x^2}{4}$ is a monotone function on $[0,1]$.

You had already found the upper bound. You can tighten the lower bound when $c<1$ so as to have $P_1$ exactly, with $$\dfrac{1-\frac{c}4}{\frac{(k-1)}4 + \frac1c} \le P_k \le \dfrac{1}{\frac {k}4 + \frac1c}$$

Henry
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  • Hi Henry, I appreciate your answer so far. One thing you mention is that while the differential equation suggests some form, you might need to adjust. At a high level, how do we know the calculus suggested answer is even remotely close? It clearly works for this particular recurrence but is it purely because of the properties of this recurrence, like the monotonicity of $x - x^2/4$? If we could use the calculus approach to "guess" a form for a lower/upper bound, then I can see one using an induction approach to optimize coefficients. Should this necessarily always work? – spektr Mar 01 '21 at 02:34
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    @spektr I doubt it always works (imagine the effect of negative $c$), but it seems a reasonable approach to find a possible useful and provable solution. Here it works well at the small end (small $c$ or large $k$) as $P_k$ and $P_{k+1}$ become relatively close, making the discreteness less important, so it becomes a matter of adjusting the bounds to work near the large end and then proving the result. – Henry Mar 01 '21 at 02:42
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    @spektr : See https://math.stackexchange.com/questions/3837335/proving-the-limiting-behavior-of-functions-containing-iterated-trigonometric and links there for other applications of this "Bernoulli trick" to recursion formulas. – Lutz Lehmann Mar 01 '21 at 09:29
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Such recursions that look like Bernoulli equations can often be solved using the "Bernoulli trick". That is, consider the corresponding negative power of $P$, $$ P_k^{-1}=P_{k-1}^{-1}\left(1-\tfrac14P_{k-1}\right)^{-1} =P_{k-1}^{-1}+\tfrac14+\tfrac1{16}P_{k-1}+\tfrac1{64}P_{k-1}^2+...\ge P_{k-1}^{-1}+\tfrac14 $$ so that $$ P_k\ge \frac{k}4+P_0^{-1}\implies P_k\le \frac{4P_0}{4+P_0k} $$

On the other side, use the observation that $P_k<P_{k-1}<...<P_1<P_0\le 1$ is strictly falling to get $$ P_k^{-1}=P_{k-1}^{-1}\left(1-\tfrac14P_{k-1}\right)^{-1} =P_{k-1}^{-1}+\tfrac14+\tfrac1{16}P_{k-1}+...\le P_{k-1}^{-1}+\tfrac14\frac{1}{1-\frac14P_0}=P_{k-1}^{-1}+\frac1{4-P_0} $$ From this one concludes $$ P_k^{-1}\le\frac{k}{4-P_0}+P_0^{-1}\implies P_k\ge \frac{4P_1}{4+P_0(k-1)} $$ So the result is the same as that of Henry, only with a more direct derivation.

Lutz Lehmann
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