Brunei Math Club

  Defining the death process Consider a population where individuals only die, and nobody is born. We now study the death process under the following assumptions. The probability that each individual dies in a short period of time $\delta t$ is $\mu \delta t+o(\delta t)$ where $\mu >0$ is the  death rate . For a population of $n$ individuals, the probability of each death is $n\mu \delta t+o(\delta t)$. We assume that the probability of multiple deaths in $\delta t$ is negligible.  Thus, the population size $N(t)$ is a random variable that only decreases. Let $$p_n(t)=Pr(N(t)=n),$$ the probability that the population size is $n$ at time $t$. Suppose the initial population size is $N(0)=n_0$, and hence the initial condition is given as $$\begin{array}{}\text{(Eq:Init)}& p_n(t)={\delta }_{n,n_0}.\end{array}$$ A sample path of a death process is shown below: Differential-difference equations Now, let's find the differential-difference equations for the death process. Fir...

  Defining the birth process  Consider a colony of bacteria that never dies. We study the following process known as  the birth process , also known as  the Yule process . The colony starts with $n_0$ cells at time $t=0$. Assume that the probability that any individual cell divides in the time interval $(t,t+\delta t)$ is proportional to $\delta t$ for small $\delta t$. Further assume that each cell division is independent of others. Let $\lambda$ be the  birth rate.  The probability of a cell division for a population of $n$ cells during $\delta t$ is $\lambda n\delta t$. We assume that the probability that two or more births take place in the time interval $\delta t$ is $o(\delta t)$. That is, it can be ignored. Consequently, the probability that no cell divides during $\delta t$ is $1-\lambda n\delta t-o(\delta t)$. Note that this process is an example of the Markov chain with states \({n_0}, {n_0 + 1}, {n_0 + 2}...

Sometimes, we need to integrate a function on a bounded closed region other than a rectangle or a region sandwiched by two curves. Even when the bounded closed region is simple, it may still be difficult to integrate a function on it. We may calculate the integral in such cases by changing the integration variables. Consider two ${\mathbb{R}}^2$ planes, the $(u,v)$ plane and $(x,y)$ plane. Suppose we have a map $\mathrm{\Phi }$ $$\mathrm{\Phi }(u,v)=(x(u,v),y(u,v))$$ defined on a neighbor of the bounded closed region $E$ in the $(u,v)$ plane. Suppose $\mathrm{\Phi }$ maps $E$ to the bounded closed region $D$ in the $(x,y)$ plane (Figure 1): $$\mathrm{\Phi }(E)=D.$$ Figure 1. Change of variables by the mapping $\mathrm{\Phi }(u,v)=(x(u,v),y(u,v))$. We further assume the following: $\mathrm{\Phi }(u,v)$ is injective. The functions $x(u,v)$ and $y(u,v)$ are of class $C^1$, and at an arbitrary point $(u_0,v_0)\in E$, we have \[\frac{\partial x}{\partial u}(u_0,v_0)\frac{\partial y}{\partial v}(u_0,v_0)-\frac{\...

Let $\mathbf{X}=(X_1,X_2,\cdots ,X_n{)}^{\mathrm{\top }}$ be a vector of random variables. We say it follows the multivariate normal (Gaussian) distribution if its density is given by $$\begin{array}{}\text{(Eq:density)}& f(\mathbf{x})=\frac{1}{\sqrt{(2\pi {)}^n|\mathrm{\Sigma }|}}\exp (-\frac{1}{2}(\mathbf{x}-\mathit{\mu }{)}^{\mathrm{\top }}{\mathrm{\Sigma }}^{-1}(\mathbf{x}-\mathit{\mu }))\end{array}$$ where $\mathit{\mu }=({\mu }_1,{\mu }_2,\cdots ,{\mu }_n{)}^{\mathrm{\top }}\in {\mathbb{R}}^n$ is a vector, $\mathrm{\Sigma }$ is a symmetric   positive definite $n\times n$ matrix, and ${\mathrm{\Sigma }}^{-1}$ and $|\mathrm{\Sigma }|$ are the inverse and determinant of $\mathrm{\Sigma }$, respectively. It turns out that $\mathit{\mu }$ and $\mathrm{\Sigma }$ are the mean vector and covariance matrix of $\mathbf{X}$, respectively. But we will not prove that here. In this post, we will show this density (Eq:density) is normalized (of course). That is, we prove that $${\int }_{{\mathbb{R}}^n}f(\mathbf{x})d\mathbf{x}=1.$$ We assume that you already know how to prove the univariate nor...

We saw the method of iterative integral for a rectangular region. We now extend the method to the case of non-rectangular regions. See also : Iterated integral on a rectangular region Consider two continuous functions $y=\phi (x)$ and  $y=\psi (x)$ on $[a,b]$ such that $\psi (x)<\phi (x)$ for all $x\in [a,b]$. We can define the following region $D$ in ${\mathbb{R}}^2$ sandwiched by these functions: $$\begin{array}{}\text{(Eq:Dsand)}& D=\{(x,y)\mid a\le x\le b,\psi (x)\le y\le \phi (x)\}.\end{array}$$ Clearly, $D$ is a closed Jordan region . Let $f(x,y)$ be a continuous function on $D$. Then, $f(x,y)$ is integrable on $D$. We can calculate the integral of $f(x,y)$ on $D$ by the method of iterated integral. See also : Multiple integral on a bounded closed set  for the explanation of closed Jordan regions. First, fix an $x\in [a,b]$, and define the univariate function $F_1(x)$ by  $$\begin{array}{}\text{(Eq:F1)}& F_1(x)={\int }_{\psi (x)}^{\phi (x)}f(x,y)dy.\end{array}$$ Lemma...

Let $\mathbf{X}=(X_1,X_2,\cdots ,X_n{)}^{\mathrm{\top }}\in {\mathbb{R}}^n$ be a vector of random variables. The covariance matrix $\mathrm{\Sigma }$ of $\mathbf{X}$ is a square ($n\times n$) matrix whose elements are covariances between the components of $\mathbf{X}$. That is, $${\mathrm{\Sigma }}_{ij}=\mathrm{Cov}(X_i,X_j)$$ where $\mathrm{Cov}(X_i,X_j)$ is the covariance between $X_i$ and $X_j$ , $i,j=1,2,\cdots ,n$:  $$\mathrm{Cov}(X_i,X_j)=\mathbb{E}[(X_i-\mathbb{E}[X_i])(X_j-\mathbb{E}[X_j])].$$ Here, $\mathbb{E}[\cdot ]$ indicates the expectation value of a random variable . Any covariance matrix has the following properties: Symmetric. That is, $$\mathrm{\Sigma }={\mathrm{\Sigma }}^{\mathrm{\top }}.$$ Positive semi-definite. That is,$$\mathrm{\forall }\mathbf{v}\in {\mathbb{R}}^n,{\mathbf{v}}^{\mathrm{\top }}\mathrm{\Sigma }\mathbf{v}\ge 0.$$ See also : Positive definite matrix (Wolfram MathWorld) The symmetry is obvious from the definition of the covariance matrix.  Now, let us prove that the covariance matrix is positive semi-...

Defining the multiple integral is one thing; calculating it is another. The iterated integral is a technique to calculate a multiple integral. Simply put, an iterated integral is a technique where we apply one-variable integration iteratively, thereby reducing a multiple integral to one-variable integrals. Here, we consider an iterated integral over a rectangular region. Let $f(x,y)$ be a continuous function on the rectangular region $D=[a,b]\times [c,d]$. Fix an arbitrary $x_0\in [a,b]$. Then, $f(x_0,y)$ can be regarded as a univariate function of $y$. $f(x_0,y)$ is continuous on $[c,d]$ so it is integrable. The integral $${\int }_c^{d}f(x_0,y)dy$$ contains $x_0$ as a parameter. Replacing $x_0$ with $x$, let us define the function $F_1(x)$ of $x$ on $[a,b]$ by $$F_1(x)={\int }_c^{d}f(x,y)dy.$$ We have the following lemma: Lemma The function $F_1(x)$ defined above is continuous on $[a,b]$. Proof . The proof will be given in another post when we prov...

In a previous post, we defined the multiple integral on a (closed) rectangular region and saw that continuous functions were integrable on such regions.  But more functions are integrable. Some functions are not necessarily defined on a rectangular region. Yet other functions may not be continuous everywhere. To describe a wider class of integrable functions, we define the notion of the "null" sets. Null sets Definition (Null set) The subset $A$ of ${\mathbb{R}}^2$ is said to be a  null set  if the following condition is satisfied: For all $\epsilon >0$, there exist a finite number of open rectangular regions  $$L_i=(a_i,b_i)\times (c_i,d_i)(a_i<b_i,c_i<d_i,i=1,2,\cdots ,r)$$  such that  $$A\subset \bigcup _{i=1}^r{L}_i$$  and  $$\sum _{i=1}^r(b_i-a_i)(d_i-c_i)<\epsilon .$$ This means that if a set is a null set if it can be "covered" by a finite number of open rectangular regions, and the total area of...

Here are some problems to test your basic knowledge and understanding. Problem 1 The absolute value $|x|$ of a real number $x$ is defined as $$|x|=\{\begin{array}{cc}x& \text{if }x\ge 0,\\ -x& \text{if }x<0.\end{array}$$ Prove the following. For any $x\in \mathbb{R}$, $-|x|\le x\le |x|$. Let $a>0$. For any $x\in \mathbb{R}$, if $-a\le x\le a$, then $|x|\le a.$  [5 marks each; 10 marks in total] Problem 2 Let $\mathbf{a}=(2,3)$, $\mathbf{b}=(1,5)$, and $\mathbf{r}=(x,y)$ be position vectors. The equation $$\mathbf{r}=s\mathbf{a}+(1-s)\mathbf{b}$$ with $0\le s\le 1$ defines the line segment between $\mathbf{a}$ and $\mathbf{b}$.  Draw the line segment defined above in the $x$-$y$ plane. Find the closest point on this line segment from the origin $\mathbf{o}=(0,0)$.  [5 marks each; 10 marks in total] Problem 3 Compute the matrix determinant ...

We extend the notion of integral to multivariate functions. For a bivariate function $f(x,y)$ on a closed region $D$, its double integral $${\iint }_{D}f(x,y)dxdy$$ is defined similarly to what we did for univariate functions, using the Riemann sums . In this and the following posts, we mainly deal with integrating bivariate (two-variable) functions (i.e., double integral). Roughly speaking, a double integral corresponds to the volume of a solid between a region on the $x$-$y$ plane and a two-variable function $z=f(x,y)$ in ${\mathbb{R}}^3$. Riemann sums The integral of multivariate functions can be defined in much the same manner as univariate functions. In the case of univariate functions, we first considered integrals on a closed interval. In the case of bivariate functions, we first consider integrals on a bounded closed region, but the region can be of any shape. We start with a rectangular region. Let $f(x,y)$ be a bounded function on the rectangular region \(D = [a,b]\t...

Consider the following problem. For any $x\in \mathbb{R}$, prove that $$-|x|\le x\le |x|.$$ In case you forgot, the absolute value of a real number $x$ is defined as follows: $$|x|=\{\begin{array}{cc}x& \text{if }x\ge 0,\\ -x& \text{if }x<0.\end{array}$$ A notable property of the absolute value is that it is always non-negative (positive or zero). Proof (wrong!): Let $x=3$. Then $|x|=3$. Thus, we have  $$-3\le 3\le 3.$$  So the proposition is true. Let $x=-3$. Then $|x|=|-3|=3$. Thus, we have $$-3\le -3\le 3.$$ So the proposition is also true. In either case, the proposition is true. Proved! ■ What is exactly wrong with this "proof"? It is that it only proves the cases when $x=3$ or $x=-3$. But we need to prove that for any real number $x$, and there are infinitely many (in fact, uncountably many) real numbers! What about $x=2.3$, or $x={\pi }^{-\sqrt{2}}$, or ...? Unless we prove the proposit...

 We now consider the differentiation of more general maps ${\mathbb{R}}^n\to {\mathbb{R}}^m$. We can import many results from the case of bivariate functions, ${\mathbb{R}}^2\to \mathbb{R}$. Multivariate functions ${\mathbb{R}}^n\to \mathbb{R}$ First, we consider the general multivariate function $y=f(x):{\mathbb{R}}^n\to \mathbb{R}$ where $x=(x_1,x_2,\cdots ,x_n)\in {\mathbb{R}}^n$. Definition (Total differentiability) Let $U$ be an open region in ${\mathbb{R}}^n$. For the function $f(x)$ on $U$ and $a=(a_1,a_2,\cdots ,a_n)\in U$, $f(x)$ is said to be (totally) differentiable at $a$ if there exist constants $m_1,m_2,\cdots ,m_n$ such that $$f(x)=f(a)+m_1(x_1-a_1)+m_2(x_2-a_2)+\cdots +m_n(x_n-a_n)+o(\Vert x-a\Vert )$$ as $x\to a$. $f(x)$ is said to be totally differentiable on $U$ if $f(x)$ is totally differentiable at all points in $U$. When $n=2$, this definition matches the definition of total differentiability of two-variabl...