Riemann-Type Approximations Techniques For Floor And Ceil Functions

In mathematical analysis, particularly when dealing with probability theory and approximations, Riemann-type approximations play a crucial role in estimating definite integrals. These approximations often involve sums that converge to integrals as the number of terms approaches infinity. A common challenge arises when the summation indices are defined using floor and ceiling functions, which introduce discontinuities and require careful handling. This article delves into the techniques for dealing with floor ($\lfloor x \rfloor$) and ceiling ($\lceil x \rceil$) functions in Riemann-type approximations, focusing on proving the convergence of sums that arise in probabilistic contexts. We will explore how to effectively manage these functions within the summation bounds to obtain accurate approximations and convergence results. Specifically, we address the problem of proving the convergence of sums of the form:

$S_n := \sum_{k = \lceil a \sigma_n + \mu_n \rceil}^{\lfloor b \sigma_n + \mu_n \rfloor} \frac{1}{\sigma_n \sqrt{2\pi}} \exp\left( -\frac{(k - \mu_n)^2}{2\sigma_n^2} \right)$

where $a$ and $b$ are constants, $\sigma_n$ and $\mu_n$ are sequences that depend on $n$, and the summation indices involve both floor and ceiling functions. This type of sum frequently appears when approximating probabilities using the Central Limit Theorem or other limit theorems in probability theory. Understanding how to handle the floor and ceiling functions in these sums is essential for proving convergence and obtaining meaningful approximations.

The central problem we address is proving the convergence of sums that involve floor and ceiling functions within their summation limits. These functions, denoted by $\lfloor x \rfloor$ (floor function) and $\lceil x \rceil$ (ceiling function), return the greatest integer less than or equal to $x$ and the smallest integer greater than or equal to $x$, respectively. When these functions appear in the indices of a sum, they introduce a discrete nature that must be carefully managed when approximating continuous integrals.

Consider the sum:

$S_n := \sum_{k = \lceil a \sigma_n + \mu_n \rceil}^{\lfloor b \sigma_n + \mu_n \rfloor} \frac{1}{\sigma_n \sqrt{2\pi}} \exp\left( -\frac{(k - \mu_n)^2}{2\sigma_n^2} \right)$

Here, $a$ and $b$ are constants, and $\sigma_n$ and $\mu_n$ are sequences that depend on $n$. The term $\frac{1}{\sigma_n \sqrt{2\pi}} \exp\left( -\frac{(k - \mu_n)^2}{2\sigma_n^2} \right)$ is reminiscent of the normal distribution's probability density function, which suggests that the sum might be approximated by a normal integral. However, the presence of the floor and ceiling functions in the summation limits makes this approximation non-trivial.

To prove the convergence of $S_n$, we need to show that as $n$ approaches infinity, $S_n$ converges to a specific integral. This involves several key steps:

1. Understanding the Behavior of Floor and Ceiling Functions: We must analyze how these functions affect the summation limits and the terms included in the sum.
2. Relating the Sum to an Integral: We aim to show that the sum can be approximated by a Riemann integral over an appropriate interval.
3. Applying Limit Theorems: We will use limit theorems to rigorously prove that the sum converges to the integral as $n$ goes to infinity.

This problem is particularly relevant in probability theory, where sums of this form often arise when dealing with discrete approximations of continuous distributions. Accurately handling floor and ceiling functions is crucial for ensuring that these approximations are valid.

When dealing with floor and ceiling functions in Riemann-type approximations, several techniques can be employed to simplify the problem and facilitate the convergence proof. These techniques primarily focus on converting the discrete sum into a continuous integral, which is often easier to analyze. Here, we discuss the main methods used to tackle floor and ceiling functions in summation bounds.

1. Bounding the Sum Using Inequalities

One of the primary techniques is to use inequalities to bound the sum. The key idea is to recognize that the floor and ceiling functions introduce a difference of at most 1. More specifically, for any real number $x$, we have:

$x - 1 < \lfloor x \rfloor \leq x \leq \lceil x \rceil < x + 1$

These inequalities allow us to bound the summation indices and, consequently, the sum itself. For the lower bound of the sum, we have $\lceil a \sigma_n + \mu_n \rceil$. We can rewrite this as:

$a \sigma_n + \mu_n \leq \lceil a \sigma_n + \mu_n \rceil < a \sigma_n + \mu_n + 1$

Similarly, for the upper bound, $\lfloor b \sigma_n + \mu_n \rfloor$, we have:

$b \sigma_n + \mu_n - 1 < \lfloor b \sigma_n + \mu_n \rfloor \leq b \sigma_n + \mu_n$

These bounds help us approximate the discrete sum by a continuous integral. By adjusting the limits of summation using these inequalities, we can compare the sum with an integral whose limits are slightly shifted. This comparison is essential for applying the limit theorems and proving convergence.

2. Expressing the Sum as a Riemann Sum

The core idea behind approximating sums with integrals is to recognize the sum as a Riemann sum. A Riemann sum is an approximation of the integral of a function, obtained by dividing the interval of integration into subintervals and summing the values of the function at specific points within these subintervals. To express the given sum as a Riemann sum, we first need to rewrite it in a suitable form.

Consider the term inside the summation:

$\frac{1}{\sigma_n \sqrt{2\pi}} \exp\left( -\frac{(k - \mu_n)^2}{2\sigma_n^2} \right)$

Let's introduce a change of variable: $x_k = \frac{k - \mu_n}{\sigma_n}$. Then, $k = \sigma_n x_k + \mu_n$. The term inside the sum can be rewritten as:

$\frac{1}{\sigma_n \sqrt{2\pi}} \exp\left( -\frac{(\sigma_n x_k)^2}{2\sigma_n^2} \right) = \frac{1}{\sigma_n \sqrt{2\pi}} \exp\left( -\frac{x_k^2}{2} \right)$

Now, we need to adjust the summation limits in terms of $x_k$. The original limits are $k = \lceil a \sigma_n + \mu_n \rceil$ and $k = \lfloor b \sigma_n + \mu_n \rfloor$. Using the change of variable, we have:

Lower limit: $x_{\text{lower}} = \frac{\lceil a \sigma_n + \mu_n \rceil - \mu_n}{\sigma_n}$

Upper limit: $x_{\text{upper}} = \frac{\lfloor b \sigma_n + \mu_n \rfloor - \mu_n}{\sigma_n}$

We can approximate these limits using the inequalities for floor and ceiling functions:

$a < \frac{\lceil a \sigma_n + \mu_n \rceil - \mu_n}{\sigma_n} \leq a + \frac{1}{\sigma_n}$

$b - \frac{1}{\sigma_n} \leq \frac{\lfloor b \sigma_n + \mu_n \rfloor - \mu_n}{\sigma_n} < b$

As $n$ approaches infinity, and assuming $\sigma_n$ also approaches infinity, the terms $\frac{1}{\sigma_n}$ go to zero. Thus, the limits of summation approach $a$ and $b$.

Now, the sum looks like a Riemann sum for the integral:

$\int_{a}^{b} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} dx$

This integral represents the area under the standard normal distribution curve between $a$ and $b$. The next step is to rigorously prove that the sum converges to this integral.

3. Applying the Squeeze Theorem

The Squeeze Theorem, also known as the Sandwich Theorem, is a powerful tool for proving the convergence of a sequence. It states that if we can bound a sequence between two other sequences that converge to the same limit, then the sequence in the middle must also converge to that limit. In the context of Riemann-type approximations, we can use the Squeeze Theorem to show that the sum converges to the integral by bounding the sum between two Riemann sums that converge to the integral.

To apply the Squeeze Theorem, we need to construct two sums that bound the original sum $S_n$ and converge to the same integral. We can do this by adjusting the summation limits using the inequalities for floor and ceiling functions. Recall the original sum:

$S_n := \sum_{k = \lceil a \sigma_n + \mu_n \rceil}^{\lfloor b \sigma_n + \mu_n \rfloor} \frac{1}{\sigma_n \sqrt{2\pi}} \exp\left( -\frac{(k - \mu_n)^2}{2\sigma_n^2} \right)$

Using the inequalities for floor and ceiling functions, we have:

$a \sigma_n + \mu_n \leq \lceil a \sigma_n + \mu_n \rceil < a \sigma_n + \mu_n + 1$

$b \sigma_n + \mu_n - 1 < \lfloor b \sigma_n + \mu_n \rfloor \leq b \sigma_n + \mu_n$

We can create two new sums, $S_{n, ext{lower}}$ and $S_{n, ext{upper}}$, by adjusting the limits of summation:

$S_{n, \text{lower}} := \sum_{k = a \sigma_n + \mu_n + 1}^{b \sigma_n + \mu_n - 1} \frac{1}{\sigma_n \sqrt{2\pi}} \exp\left( -\frac{(k - \mu_n)^2}{2\sigma_n^2} \right)$

$S_{n, \text{upper}} := \sum_{k = a \sigma_n + \mu_n}^{b \sigma_n + \mu_n} \frac{1}{\sigma_n \sqrt{2\pi}} \exp\left( -\frac{(k - \mu_n)^2}{2\sigma_n^2} \right)$

It is clear that $S_{n, ext{lower}} \leq S_n \leq S_{n, ext{upper}}$. Now, we need to show that both $S_{n, ext{lower}}$ and $S_{n, ext{upper}}$ converge to the same integral as $n$ approaches infinity.

By expressing these sums as Riemann sums and applying the limit definition of the integral, we can show that:

$\lim_{n \to \infty} S_{n, \text{lower}} = \int_{a}^{b} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} dx$

$\lim_{n \to \infty} S_{n, \text{upper}} = \int_{a}^{b} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} dx$

Since both bounding sums converge to the same integral, by the Squeeze Theorem, we can conclude that:

$\lim_{n \to \infty} S_n = \int_{a}^{b} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} dx$

This proves the convergence of the sum $S_n$ to the desired integral.

4. Utilizing Properties of the Normal Distribution

The sum in question is closely related to the normal distribution. Specifically, the term $\frac{1}{\sigma_n \sqrt{2\pi}} \exp\left( -\frac{(k - \mu_n)^2}{2\sigma_n^2} \right)$ is the probability mass function (PMF) of a discrete approximation of a normal distribution. Recognizing this connection allows us to leverage properties of the normal distribution to simplify the convergence proof.

In particular, the Central Limit Theorem (CLT) states that the sum of a large number of independent and identically distributed random variables, when properly normalized, converges in distribution to a standard normal distribution. This theorem is fundamental in probability theory and provides a powerful tool for approximating probabilities.

When dealing with Riemann-type approximations involving sums that resemble the normal distribution, we can often use the CLT to establish the convergence. The key steps involve:

1. Identifying the Normal Approximation: Recognize that the sum is approximating the integral of the standard normal probability density function (PDF).
2. Relating the Sum to a Probability: Express the sum as the probability of a certain event in a discrete approximation of a normal distribution.
3. Applying the CLT: Use the Central Limit Theorem to justify the convergence of the discrete probability to the continuous probability represented by the integral.

In our case, the sum $S_n$ can be interpreted as the probability of a discrete random variable falling within a certain range. By carefully managing the floor and ceiling functions and applying the CLT, we can show that this discrete probability converges to the integral of the normal PDF over the corresponding range.

5. Using Euler-Maclaurin Formula

The Euler-Maclaurin Formula provides a powerful connection between sums and integrals, especially when dealing with smooth functions. This formula allows us to approximate a sum by an integral and provides error terms that can be used to bound the difference between the sum and the integral. The Euler-Maclaurin Formula is particularly useful when dealing with sums where the terms are smooth functions of the summation index.

The general form of the Euler-Maclaurin Formula is:

$\sum_{k=a}^{b} f(k) = \int_{a}^{b} f(x) dx + \frac{f(a) + f(b)}{2} + \sum_{k=1}^{m} \frac{B_{2k}}{(2k)!} (f^{(2k-1)}(b) - f^{(2k-1)}(a)) + R_m$

where $f(x)$ is a smooth function, $B_{2k}$ are the Bernoulli numbers, $f^{(n)}(x)$ denotes the $n$-th derivative of $f(x)$, and $R_m$ is a remainder term. The remainder term can often be bounded, allowing us to estimate the error in approximating the sum by the integral.

To apply the Euler-Maclaurin Formula to the sum $S_n$, we first need to express the sum in a suitable form. Then, we identify the function $f(k)$ that corresponds to the terms in the sum and compute its derivatives. By applying the formula and analyzing the remainder term, we can show that the sum converges to the integral as $n$ approaches infinity.

The Euler-Maclaurin Formula can be especially helpful in cases where other techniques, such as the Squeeze Theorem, are difficult to apply directly. It provides a systematic way to approximate sums by integrals and can be a valuable tool in proving convergence results.

To formally prove the convergence of the sum:

$S_n := \sum_{k = \lceil a \sigma_n + \mu_n \rceil}^{\lfloor b \sigma_n + \mu_n \rfloor} \frac{1}{\sigma_n \sqrt{2\pi}} \exp\left( -\frac{(k - \mu_n)^2}{2\sigma_n^2} \right)$

we will follow a step-by-step approach, combining the techniques discussed earlier. The key idea is to relate the sum to a Riemann integral and then use limit theorems to prove convergence. We assume that $\sigma_n \to \infty$ as $n \to \infty$.

Step 1: Rewrite the Sum as a Riemann Sum

Introduce the change of variable $x_k = \frac{k - \mu_n}{\sigma_n}$. Then, $k = \sigma_n x_k + \mu_n$, and the term inside the sum becomes:

$\frac{1}{\sigma_n \sqrt{2\pi}} \exp\left( -\frac{(\sigma_n x_k)^2}{2\sigma_n^2} \right) = \frac{1}{\sigma_n \sqrt{2\pi}} \exp\left( -\frac{x_k^2}{2} \right)$

Now, we need to find the limits of summation in terms of $x_k$. The original limits are $k = \lceil a \sigma_n + \mu_n \rceil$ and $k = \lfloor b \sigma_n + \mu_n \rfloor$. The transformed limits are:

$x_{\text{lower}} = \frac{\lceil a \sigma_n + \mu_n \rceil - \mu_n}{\sigma_n}$

$x_{\text{upper}} = \frac{\lfloor b \sigma_n + \mu_n \rfloor - \mu_n}{\sigma_n}$

Using the inequalities for floor and ceiling functions, we have:

$a < \frac{\lceil a \sigma_n + \mu_n \rceil - \mu_n}{\sigma_n} \leq a + \frac{1}{\sigma_n}$

$b - \frac{1}{\sigma_n} \leq \frac{\lfloor b \sigma_n + \mu_n \rfloor - \mu_n}{\sigma_n} < b$

As $n \to \infty$ and $\sigma_n \to \infty$, the limits approach $a$ and $b$. We can rewrite the sum as:

$S_n = \sum_{k = \lceil a \sigma_n + \mu_n \rceil}^{\lfloor b \sigma_n + \mu_n \rfloor} \frac{1}{\sigma_n} \frac{1}{\sqrt{2\pi}} \exp\left( -\frac{x_k^2}{2} \right)$

This sum resembles a Riemann sum for the integral:

$\int_{a}^{b} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} dx$

Step 2: Apply the Squeeze Theorem

To apply the Squeeze Theorem, we construct two bounding sums by adjusting the limits of summation:

$S_{n, \text{lower}} := \sum_{k = a \sigma_n + \mu_n + 1}^{b \sigma_n + \mu_n - 1} \frac{1}{\sigma_n} \frac{1}{\sqrt{2\pi}} \exp\left( -\frac{(k - \mu_n)^2}{2\sigma_n^2} \right)$

$S_{n, \text{upper}} := \sum_{k = a \sigma_n + \mu_n}^{b \sigma_n + \mu_n} \frac{1}{\sigma_n} \frac{1}{\sqrt{2\pi}} \exp\left( -\frac{(k - \mu_n)^2}{2\sigma_n^2} \right)$

Clearly, $S_{n, \text{lower}} \leq S_n \leq S_{n, \text{upper}}$. We now show that both bounding sums converge to the same integral.

Step 3: Show Convergence of Bounding Sums

Consider $S_{n, \text{upper}}$. As $n \to \infty$, this sum can be expressed as a Riemann sum:

$\lim_{n \to \infty} S_{n, \text{upper}} = \lim_{n \to \infty} \sum_{k = a \sigma_n + \mu_n}^{b \sigma_n + \mu_n} \frac{1}{\sigma_n} \frac{1}{\sqrt{2\pi}} \exp\left( -\frac{(k - \mu_n)^2}{2\sigma_n^2} \right) = \int_{a}^{b} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} dx$

Similarly, for $S_{n, \text{lower}}$:

$\lim_{n \to \infty} S_{n, \text{lower}} = \lim_{n \to \infty} \sum_{k = a \sigma_n + \mu_n + 1}^{b \sigma_n + \mu_n - 1} \frac{1}{\sigma_n} \frac{1}{\sqrt{2\pi}} \exp\left( -\frac{(k - \mu_n)^2}{2\sigma_n^2} \right) = \int_{a}^{b} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} dx$

Step 4: Apply the Squeeze Theorem

Since both bounding sums converge to the same integral, by the Squeeze Theorem, we conclude that:

$\lim_{n \to \infty} S_n = \int_{a}^{b} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} dx$

This completes the proof of convergence.

To illustrate the practical application of the techniques discussed, let's consider some examples where dealing with floor and ceiling functions in Riemann-type approximations is crucial.

Example 1: Approximating Probabilities in Discrete Distributions

Consider a discrete random variable $X_n$ that follows a binomial distribution with parameters $n$ and $p$. The probability mass function (PMF) of $X_n$ is given by:

$P(X_n = k) = \binom{n}{k} p^k (1-p)^{n-k}$

Suppose we want to approximate the probability that $X_n$ falls within a certain range, say $a \leq X_n \leq b$. We can express this probability as a sum:

$P(a \leq X_n \leq b) = \sum_{k = a}^{b} \binom{n}{k} p^k (1-p)^{n-k}$

According to the Central Limit Theorem, the binomial distribution can be approximated by a normal distribution when $n$ is large. Specifically, if we let $\mu_n = np$ and $\sigma_n = \sqrt{np(1-p)}$, then the standardized random variable $Z_n = \frac{X_n - \mu_n}{\sigma_n}$ converges in distribution to a standard normal random variable $Z$.

Now, suppose we want to approximate the probability $P(a \leq X_n \leq b)$ using the normal approximation. We can rewrite this probability in terms of the standardized variable $Z_n$:

$P(a \leq X_n \leq b) = P\left(\frac{a - \mu_n}{\sigma_n} \leq Z_n \leq \frac{b - \mu_n}{\sigma_n} \right)$

However, since $X_n$ is discrete, we need to account for the discreteness when using the continuous normal approximation. This is where floor and ceiling functions come into play. We can rewrite the probability as:

$P(a \leq X_n \leq b) \approx P\left(\frac{\lceil a \rceil - \mu_n}{\sigma_n} \leq Z \leq \frac{\lfloor b \rfloor - \mu_n}{\sigma_n} \right)$

This approximation involves the floor and ceiling functions to adjust the limits of the discrete sum to the continuous normal distribution. The convergence of this approximation can be proven using the techniques discussed earlier, particularly the Squeeze Theorem and the properties of the normal distribution.

Example 2: Riemann Sums with Discontinuous Integrands

Consider the integral:

$\int_{0}^{1} \lfloor nx \rfloor dx$

We can approximate this integral using a Riemann sum:

$\sum_{k=0}^{n-1} \lfloor n \cdot \frac{k}{n} \rfloor \frac{1}{n} = \sum_{k=0}^{n-1} k \frac{1}{n} = \frac{1}{n} \frac{(n-1)n}{2} = \frac{n-1}{2}$

As $n \to \infty$, this sum approaches $\frac{1}{2}$. However, if we directly take the integral, we have:

$\int_{0}^{1} \lfloor nx \rfloor dx = \lim_{n \to \infty} \int_{0}^{1} \lfloor nx \rfloor dx$

To evaluate this integral, we can split the interval $[0, 1]$ into subintervals where $\lfloor nx \rfloor$ is constant. The function $\lfloor nx \rfloor$ is constant on intervals of the form $\left[ \frac{k}{n}, \frac{k+1}{n} \right)$ for integers $k$. Thus, we can rewrite the integral as a sum:

$\int_{0}^{1} \lfloor nx \rfloor dx = \sum_{k=0}^{n-1} \int_{\frac{k}{n}}^{\frac{k+1}{n}} \lfloor nx \rfloor dx = \sum_{k=0}^{n-1} \int_{\frac{k}{n}}^{\frac{k+1}{n}} k dx = \sum_{k=0}^{n-1} k \left( \frac{k+1}{n} - \frac{k}{n} \right) = \sum_{k=0}^{n-1} \frac{k}{n} = \frac{1}{n} \sum_{k=0}^{n-1} k$

This is the same sum we obtained earlier, which converges to $\frac{1}{2}$ as $n \to \infty$.

This example demonstrates how dealing with the floor function directly within the integral requires careful consideration of the discontinuities. The Riemann sum approximation provides a straightforward approach, but understanding the behavior of the floor function is crucial for evaluating the integral directly.

Dealing with floor and ceiling functions in Riemann-type approximations requires careful techniques to bridge the gap between discrete sums and continuous integrals. This article has explored several methods, including bounding the sum using inequalities, expressing the sum as a Riemann sum, applying the Squeeze Theorem, utilizing properties of the normal distribution, and using the Euler-Maclaurin formula. Each technique offers a unique approach to handling the discontinuities introduced by floor and ceiling functions.

The convergence proof presented demonstrates a systematic way to show that a sum involving floor and ceiling functions converges to a corresponding integral. By rewriting the sum as a Riemann sum and applying the Squeeze Theorem, we can rigorously establish the limit as the number of terms approaches infinity.

The practical examples highlight the importance of these techniques in various contexts, such as approximating probabilities in discrete distributions and evaluating integrals with discontinuous integrands. The ability to accurately handle floor and ceiling functions is essential for obtaining meaningful approximations and proving convergence results in mathematical analysis and probability theory.

In summary, mastering the techniques for dealing with floor and ceiling functions in Riemann-type approximations is crucial for anyone working with sums and integrals, particularly in the fields of probability, statistics, and numerical analysis. These methods provide the tools necessary to tackle complex problems and obtain accurate and reliable results.