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Birth process

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  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}...
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Computing integrals (3): Integration by parts

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 Sometimes, we may simplify integration by using the product rule of differentiation. This technique is called integration by parts. Theorem (Integration by parts) Let \(f(x)\) and \(g(x)\) be differentiable functions on an open interval \(I\). Then,  \(\int f(x)g'(x)dx = f(x)g(x) - \int f'(x)g(x)dx\); For any \(a, b \in I\), \[\int_a^bf(x)g'(x)dx = \left[f(x)g(x)\right]_a^b - \int_a^bf'(x)g(x)dx.\] Proof . By the product rule, \[[f(x)g(x)]' = f'(x)g(x) + f(x)g'(x)\] so \[f(x)g'(x) = [f(x)g(x)]' - f'(x)g(x).\] By integrating both sides, we have the desired results. ■ Example . Let us find \(\int x\cosh x dx\). \[ \begin{eqnarray*} \int x\cosh x dx &=& \int x(\sinh x)'dx \\ &=& x \sinh x - \int 1 \cdot \sinh x dx\\ &=& x \sinh x - \cosh x + C. \end{eqnarray*} \] Example (eg:recur) . Let us study how we can compute \[I_n = \int \frac{dx}{(x^2 + 1)^n}\] for \(n\in \mathbb{N}\). Note \[I_{n} = \int \fr...
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Multivariate normal distribution is normalized (of course): A proof

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Let \(\mathbf{X} = (X_1, X_2, \cdots, X_n)^\top\) be a vector of random variables. We say it follows the multivariate normal (Gaussian) distribution if its density is given by \[f(\mathbf{x}) = \frac{1}{\sqrt{(2\pi)^n|\Sigma|}}\exp\left(-\frac{1}{2}(\mathbf{x} - \boldsymbol{\mu})^\top\Sigma^{-1}(\mathbf{x} - \boldsymbol{\mu})\right)\tag{Eq:density}\] where \(\boldsymbol{\mu} = (\mu_1, \mu_2, \cdots, \mu_n)^\top \in \mathbb{R}^n\) is a vector, \(\Sigma\) is a symmetric   positive definite \(n\times n\) matrix, and \(\Sigma^{-1}\) and \(|\Sigma|\) are the inverse and determinant of \(\Sigma\), respectively. It turns out that \(\boldsymbol{\mu}\) and \(\Sigma\) 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...
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