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One-sided concentration bound for independent median trials
Statement
Let be an integer, let be independent Bernoulli random variables on a finite probability space with , and put . For every , Consequently, let be odd and let be independent finite-valued real estimates on a finite probability space. If is an interval and for every , then their median (the middle order statistic) lies in except with probability at most
Facts & Assumptions
Given: A finite probability space, an integer , and an independent Bernoulli family with means . For the median consequence, also the independent finite-valued estimates , the interval , and for each .
Bernoulli random variables and binomial random variables as sums of independent Bernoulli trials: a Bernoulli variable takes values with probabilities .
Independent random elements: mutual independence means the generated event sigma-algebras factor on finite intersections.
Expectation factors over a finite product of mutually independent random variables: expectations of finite products of independent finite-valued real variables factor.
Markov's inequality on a finite probability space: for nonnegative and , .
The real exponential function and the number by a power series fixes and ; The exponential is positive and satisfies gives ; The exponential addition formula gives ; The exponential function is strictly increasing makes order preserving; and The natural logarithm as the inverse of the exponential function defines as its inverse.
The exponential function is smooth and , The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t, The chain rule, in one line from Carathéodory: if is differentiable at and is differentiable at , then is differentiable at with , and Sums, scalar multiples, products and quotients: , , , and when give the derivative rules for the positive exponential, logarithm, sums, products, and quotients used below.
The mean value theorem, as the case of Cauchy's: for continuous on with and differentiable on there is with : if a function is continuous on and differentiable on , then its endpoint increment is its derivative at some interior point times .
Expectation of a real random variable on a finite probability space: the expectation of a finite-valued variable is the finite weighted sum of its values, including the Bernoulli formula.
Linearity, monotonicity, and the modulus bound for expectation: expectation is linear for the finite-valued real random variables below.
A function differentiable at is continuous at : a function differentiable at a point is continuous there.
Expectation agrees with the published finite weighted sum identifies general expectations of finite random variables on this probability space with the finite weighted sums in [F8].
Proof
For one Bernoulli variable , [F1] and [F8] give . Define by its two possible values and [F5], and put . Set . Differentiating by [F6] gives and , including and .
At the desired bound is equality. For , put on . By step 1.1, . The functions and are continuous by [F10], so the hypotheses of [F7] apply. For any , [F7] applied to on gives for some ; since , this shows . Applying [F7] to on then gives for some . As , . Therefore for every .
By [F11], the general expectations in [F3] and [F9] agree with the finite weighted sums in [F8]. Thus [F8]–[F9] give and . For , put . Since each is finite-valued, every event defined by is a union of level sets of , so ; the independence in [F2] therefore makes the independent. The exponential addition law [F5] and finite product factorization [F3] now give by step 2.1.
For , choose and apply [F4] to the nonnegative variable at threshold . Step 3.1 gives . For , the probability is at most one, equal to the stated bound.
Let indicate the event . Each is Bernoulli with mean at most by [F1], [F8], and the interval-success hypothesis. Each is a function of the finite-valued , so every event defined by is a union of level sets of ; hence and [F2] makes the independent. By [F11], [F9] applies to the same finite weighted-sum expectations as [F8], so . If the odd median is outside , at least estimates lie on one side of and fail, so and . Apply step 4.1 with this to get the stated median bound.
Depends on
- Finite probability spaces, outcome weights, events, and event probabilities
- Bernoulli random variables and binomial random variables as sums of independent Bernoulli trials
- Independent random elements
- Expectation factors over a finite product of mutually independent random variables
- Markov's inequality on a finite probability space
- The real exponential function and the number $e$ by a power series
- The exponential addition formula $\exp(x+y)=\exp(x)\exp(y)$
- The exponential is positive and satisfies $\exp(-x)=1/\exp(x)$
- The exponential function is strictly increasing
- Expectation of a real random variable on a finite probability space
- Linearity, monotonicity, and the modulus bound for expectation
- A function differentiable at $c$ is continuous at $c$
- The natural logarithm as the inverse of the exponential function
- The exponential function is smooth and $(\exp)'=\exp$
- The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t
- The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g(c)$, then $f \circ g$ is differentiable at $c$ with $(f \circ g)'(c) = f'(g(c))\,g'(c)$
- Sums, scalar multiples, products and quotients: $(f+g)'(c) = f'(c) + g'(c)$, $(\alpha f)'(c) = \alpha f'(c)$, $(fg)'(c) = f'(c)g(c) + f(c)g'(c)$, and $(f/g)'(c) = \bigl(f'(c)g(c) - f(c)g'(c)\bigr)/g(c)^{2}$ when $g(c) \ne 0$
- The mean value theorem, as the case $g(x) = x$ of Cauchy's: for $f$ continuous on $[a,b]$ with $a < b$ and differentiable on $(a,b)$ there is $c \in (a,b)$ with $f(b) - f(a) = f'(c)(b-a)$
- Expectation agrees with the published finite weighted sum
Used by
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Sources
- Ashley Montanaro, Quantum speedup of Monte Carlo methods, Lemma 1 and Theorems 3 and 5 (standard reference, not scraped)
- Andrew M. Childs, Lecture Notes on Quantum Algorithms, Section 19.3 (standard reference, not scraped)