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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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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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Any covariance matrix is positive semi-definite: A proof

Let \(\mathbf{X} = (X_1, X_2, \cdots, X_n)^\top \in \mathbb{R}^n\) be a vector of random variables. The covariance matrix \(\Sigma\) of \(\mathbf{X}\) is a square (\(n\times n\)) matrix whose elements are covariances between the components of \(\mathbf{X}\). That is, \[\Sigma_{ij} = \mathrm{Cov}(X_i,X_j)\] 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, \[\Sigma = \Sigma^\top.\] Positive semi-definite. That is,\[\forall \mathbf{v} \in \mathbb{R}^n, \mathbf{v}^\top\Sigma\mathbf{v} \geq 0.\] 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-...
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