linear transformation of normal distribution
Find the distribution function and probability density function of the following variables. When the transformed variable \(Y\) has a discrete distribution, the probability density function of \(Y\) can be computed using basic rules of probability. The Exponential distribution is studied in more detail in the chapter on Poisson Processes. If you have run a histogram to check your data and it looks like any of the pictures below, you can simply apply the given transformation to each participant . The multivariate version of this result has a simple and elegant form when the linear transformation is expressed in matrix-vector form. Given our previous result, the one for cylindrical coordinates should come as no surprise. This is known as the change of variables formula. Clearly we can simulate a value of the Cauchy distribution by \( X = \tan\left(-\frac{\pi}{2} + \pi U\right) \) where \( U \) is a random number. First, for \( (x, y) \in \R^2 \), let \( (r, \theta) \) denote the standard polar coordinates corresponding to the Cartesian coordinates \((x, y)\), so that \( r \in [0, \infty) \) is the radial distance and \( \theta \in [0, 2 \pi) \) is the polar angle. Set \(k = 1\) (this gives the minimum \(U\)). Recall that if \((X_1, X_2, X_3)\) is a sequence of independent random variables, each with the standard uniform distribution, then \(f\), \(f^{*2}\), and \(f^{*3}\) are the probability density functions of \(X_1\), \(X_1 + X_2\), and \(X_1 + X_2 + X_3\), respectively. Proposition Let be a multivariate normal random vector with mean and covariance matrix . From part (b) it follows that if \(Y\) and \(Z\) are independent variables, and that \(Y\) has the binomial distribution with parameters \(n \in \N\) and \(p \in [0, 1]\) while \(Z\) has the binomial distribution with parameter \(m \in \N\) and \(p\), then \(Y + Z\) has the binomial distribution with parameter \(m + n\) and \(p\). An ace-six flat die is a standard die in which faces 1 and 6 occur with probability \(\frac{1}{4}\) each and the other faces with probability \(\frac{1}{8}\) each. \(X\) is uniformly distributed on the interval \([-1, 3]\). Find the probability density function of \(Z = X + Y\) in each of the following cases. Note that the PDF \( g \) of \( \bs Y \) is constant on \( T \). (1) (1) x N ( , ). The distribution of \( R \) is the (standard) Rayleigh distribution, and is named for John William Strutt, Lord Rayleigh. In the last exercise, you can see the behavior predicted by the central limit theorem beginning to emerge. If \( a, \, b \in (0, \infty) \) then \(f_a * f_b = f_{a+b}\). If \(B \subseteq T\) then \[\P(\bs Y \in B) = \P[r(\bs X) \in B] = \P[\bs X \in r^{-1}(B)] = \int_{r^{-1}(B)} f(\bs x) \, d\bs x\] Using the change of variables \(\bs x = r^{-1}(\bs y)\), \(d\bs x = \left|\det \left( \frac{d \bs x}{d \bs y} \right)\right|\, d\bs y\) we have \[\P(\bs Y \in B) = \int_B f[r^{-1}(\bs y)] \left|\det \left( \frac{d \bs x}{d \bs y} \right)\right|\, d \bs y\] So it follows that \(g\) defined in the theorem is a PDF for \(\bs Y\). This follows from part (a) by taking derivatives with respect to \( y \). \( G(y) = \P(Y \le y) = \P[r(X) \le y] = \P\left[X \le r^{-1}(y)\right] = F\left[r^{-1}(y)\right] \) for \( y \in T \). Find the probability density function of each of the following: Suppose that the grades on a test are described by the random variable \( Y = 100 X \) where \( X \) has the beta distribution with probability density function \( f \) given by \( f(x) = 12 x (1 - x)^2 \) for \( 0 \le x \le 1 \). The Rayleigh distribution in the last exercise has CDF \( H(r) = 1 - e^{-\frac{1}{2} r^2} \) for \( 0 \le r \lt \infty \), and hence quantle function \( H^{-1}(p) = \sqrt{-2 \ln(1 - p)} \) for \( 0 \le p \lt 1 \). The Jacobian of the inverse transformation is the constant function \(\det (\bs B^{-1}) = 1 / \det(\bs B)\). Find the probability density function of. By far the most important special case occurs when \(X\) and \(Y\) are independent. Suppose that \(T\) has the gamma distribution with shape parameter \(n \in \N_+\). More generally, if \((X_1, X_2, \ldots, X_n)\) is a sequence of independent random variables, each with the standard uniform distribution, then the distribution of \(\sum_{i=1}^n X_i\) (which has probability density function \(f^{*n}\)) is known as the Irwin-Hall distribution with parameter \(n\). The problem is my data appears to be normally distributed, i.e., there are a lot of 0.999943 and 0.99902 values. In the order statistic experiment, select the uniform distribution. }, \quad 0 \le t \lt \infty \] With a positive integer shape parameter, as we have here, it is also referred to as the Erlang distribution, named for Agner Erlang. In statistical terms, \( \bs X \) corresponds to sampling from the common distribution.By convention, \( Y_0 = 0 \), so naturally we take \( f^{*0} = \delta \). As in the discrete case, the formula in (4) not much help, and it's usually better to work each problem from scratch. Find the probability density function of \(Z^2\) and sketch the graph. . In a normal distribution, data is symmetrically distributed with no skew. Using the change of variables theorem, the joint PDF of \( (U, V) \) is \( (u, v) \mapsto f(u, v / u)|1 /|u| \). We have seen this derivation before. \(g(u, v, w) = \frac{1}{2}\) for \((u, v, w)\) in the rectangular region \(T \subset \R^3\) with vertices \(\{(0,0,0), (1,0,1), (1,1,0), (0,1,1), (2,1,1), (1,1,2), (1,2,1), (2,2,2)\}\). The result now follows from the multivariate change of variables theorem. Suppose that \((X_1, X_2, \ldots, X_n)\) is a sequence of independent real-valued random variables, with common distribution function \(F\). Then \(X = F^{-1}(U)\) has distribution function \(F\). Using the theorem on quotient above, the PDF \( f \) of \( T \) is given by \[f(t) = \int_{-\infty}^\infty \phi(x) \phi(t x) |x| dx = \frac{1}{2 \pi} \int_{-\infty}^\infty e^{-(1 + t^2) x^2/2} |x| dx, \quad t \in \R\] Using symmetry and a simple substitution, \[ f(t) = \frac{1}{\pi} \int_0^\infty x e^{-(1 + t^2) x^2/2} dx = \frac{1}{\pi (1 + t^2)}, \quad t \in \R \]. Suppose that \((X, Y)\) probability density function \(f\). \(g(v) = \frac{1}{\sqrt{2 \pi v}} e^{-\frac{1}{2} v}\) for \( 0 \lt v \lt \infty\). Suppose also \( Y = r(X) \) where \( r \) is a differentiable function from \( S \) onto \( T \subseteq \R^n \). \(X = -\frac{1}{r} \ln(1 - U)\) where \(U\) is a random number. Random variable \( V = X Y \) has probability density function \[ v \mapsto \int_{-\infty}^\infty f(x, v / x) \frac{1}{|x|} dx \], Random variable \( W = Y / X \) has probability density function \[ w \mapsto \int_{-\infty}^\infty f(x, w x) |x| dx \], We have the transformation \( u = x \), \( v = x y\) and so the inverse transformation is \( x = u \), \( y = v / u\). Then we can find a matrix A such that T(x)=Ax. Find the probability density function of \((U, V, W) = (X + Y, Y + Z, X + Z)\). Recall that \( F^\prime = f \). Since \(1 - U\) is also a random number, a simpler solution is \(X = -\frac{1}{r} \ln U\). }, \quad n \in \N \] This distribution is named for Simeon Poisson and is widely used to model the number of random points in a region of time or space; the parameter \(t\) is proportional to the size of the regtion. However, the last exercise points the way to an alternative method of simulation. Convolution can be generalized to sums of independent variables that are not of the same type, but this generalization is usually done in terms of distribution functions rather than probability density functions. The first image below shows the graph of the distribution function of a rather complicated mixed distribution, represented in blue on the horizontal axis. . Find the probability density function of \(X = \ln T\). Now we can prove that every linear transformation is a matrix transformation, and we will show how to compute the matrix. . Show how to simulate a pair of independent, standard normal variables with a pair of random numbers. Next, for \( (x, y, z) \in \R^3 \), let \( (r, \theta, z) \) denote the standard cylindrical coordinates, so that \( (r, \theta) \) are the standard polar coordinates of \( (x, y) \) as above, and coordinate \( z \) is left unchanged. Suppose now that we have a random variable \(X\) for the experiment, taking values in a set \(S\), and a function \(r\) from \( S \) into another set \( T \). (z - x)!} Hence the PDF of W is \[ w \mapsto \int_{-\infty}^\infty f(u, u w) |u| du \], Random variable \( V = X Y \) has probability density function \[ v \mapsto \int_{-\infty}^\infty g(x) h(v / x) \frac{1}{|x|} dx \], Random variable \( W = Y / X \) has probability density function \[ w \mapsto \int_{-\infty}^\infty g(x) h(w x) |x| dx \]. In the context of the Poisson model, part (a) means that the \( n \)th arrival time is the sum of the \( n \) independent interarrival times, which have a common exponential distribution. Transforming data to normal distribution in R. I've imported some data from Excel, and I'd like to use the lm function to create a linear regression model of the data. For \(i \in \N_+\), the probability density function \(f\) of the trial variable \(X_i\) is \(f(x) = p^x (1 - p)^{1 - x}\) for \(x \in \{0, 1\}\). \(f(u) = \left(1 - \frac{u-1}{6}\right)^n - \left(1 - \frac{u}{6}\right)^n, \quad u \in \{1, 2, 3, 4, 5, 6\}\), \(g(v) = \left(\frac{v}{6}\right)^n - \left(\frac{v - 1}{6}\right)^n, \quad v \in \{1, 2, 3, 4, 5, 6\}\). Find the probability density function of each of the following random variables: In the previous exercise, \(V\) also has a Pareto distribution but with parameter \(\frac{a}{2}\); \(Y\) has the beta distribution with parameters \(a\) and \(b = 1\); and \(Z\) has the exponential distribution with rate parameter \(a\). If \( A \subseteq (0, \infty) \) then \[ \P\left[\left|X\right| \in A, \sgn(X) = 1\right] = \P(X \in A) = \int_A f(x) \, dx = \frac{1}{2} \int_A 2 \, f(x) \, dx = \P[\sgn(X) = 1] \P\left(\left|X\right| \in A\right) \], The first die is standard and fair, and the second is ace-six flat. The family of beta distributions and the family of Pareto distributions are studied in more detail in the chapter on Special Distributions. Hence the following result is an immediate consequence of the change of variables theorem (8): Suppose that \( (X, Y, Z) \) has a continuous distribution on \( \R^3 \) with probability density function \( f \), and that \( (R, \Theta, \Phi) \) are the spherical coordinates of \( (X, Y, Z) \). Find the probability density function of the following variables: Let \(U\) denote the minimum score and \(V\) the maximum score. If \( (X, Y) \) has a discrete distribution then \(Z = X + Y\) has a discrete distribution with probability density function \(u\) given by \[ u(z) = \sum_{x \in D_z} f(x, z - x), \quad z \in T \], If \( (X, Y) \) has a continuous distribution then \(Z = X + Y\) has a continuous distribution with probability density function \(u\) given by \[ u(z) = \int_{D_z} f(x, z - x) \, dx, \quad z \in T \], \( \P(Z = z) = \P\left(X = x, Y = z - x \text{ for some } x \in D_z\right) = \sum_{x \in D_z} f(x, z - x) \), For \( A \subseteq T \), let \( C = \{(u, v) \in R \times S: u + v \in A\} \). Graph \( f \), \( f^{*2} \), and \( f^{*3} \)on the same set of axes. Find the probability density function of each of the following random variables: Note that the distributions in the previous exercise are geometric distributions on \(\N\) and on \(\N_+\), respectively. . The Cauchy distribution is studied in detail in the chapter on Special Distributions. Recall again that \( F^\prime = f \). In the classical linear model, normality is usually required. Find the probability density function of each of the following: Random variables \(X\), \(U\), and \(V\) in the previous exercise have beta distributions, the same family of distributions that we saw in the exercise above for the minimum and maximum of independent standard uniform variables. Note that the minimum \(U\) in part (a) has the exponential distribution with parameter \(r_1 + r_2 + \cdots + r_n\). To check if the data is normally distributed I've used qqplot and qqline . With \(n = 4\), run the simulation 1000 times and note the agreement between the empirical density function and the probability density function. Here is my code from torch.distributions.normal import Normal from torch. I want to compute the KL divergence between a Gaussian mixture distribution and a normal distribution using sampling method. Recall that the sign function on \( \R \) (not to be confused, of course, with the sine function) is defined as follows: \[ \sgn(x) = \begin{cases} -1, & x \lt 0 \\ 0, & x = 0 \\ 1, & x \gt 0 \end{cases} \], Suppose again that \( X \) has a continuous distribution on \( \R \) with distribution function \( F \) and probability density function \( f \), and suppose in addition that the distribution of \( X \) is symmetric about 0. For our next discussion, we will consider transformations that correspond to common distance-angle based coordinate systemspolar coordinates in the plane, and cylindrical and spherical coordinates in 3-dimensional space. Then \(\bs Y\) is uniformly distributed on \(T = \{\bs a + \bs B \bs x: \bs x \in S\}\). Moreover, this type of transformation leads to simple applications of the change of variable theorems. Similarly, \(V\) is the lifetime of the parallel system which operates if and only if at least one component is operating. Suppose that \(\bs X = (X_1, X_2, \ldots)\) is a sequence of independent and identically distributed real-valued random variables, with common probability density function \(f\). The dice are both fair, but the first die has faces labeled 1, 2, 2, 3, 3, 4 and the second die has faces labeled 1, 3, 4, 5, 6, 8. f Z ( x) = 3 f Y ( x) 4 where f Z and f Y are the pdfs. Since \( X \) has a continuous distribution, \[ \P(U \ge u) = \P[F(X) \ge u] = \P[X \ge F^{-1}(u)] = 1 - F[F^{-1}(u)] = 1 - u \] Hence \( U \) is uniformly distributed on \( (0, 1) \). However, when dealing with the assumptions of linear regression, you can consider transformations of . I have an array of about 1000 floats, all between 0 and 1. Suppose that \(T\) has the exponential distribution with rate parameter \(r \in (0, \infty)\). Suppose that \(X\) has a continuous distribution on \(\R\) with distribution function \(F\) and probability density function \(f\). The result now follows from the change of variables theorem. Convolution (either discrete or continuous) satisfies the following properties, where \(f\), \(g\), and \(h\) are probability density functions of the same type. e^{t-s} \, ds = e^{-t} \int_0^t \frac{s^{n-1}}{(n - 1)!} When plotted on a graph, the data follows a bell shape, with most values clustering around a central region and tapering off as they go further away from the center. When V and W are finite dimensional, a general linear transformation can Algebra Examples. Let \(\bs Y = \bs a + \bs B \bs X\) where \(\bs a \in \R^n\) and \(\bs B\) is an invertible \(n \times n\) matrix. The Poisson distribution is studied in detail in the chapter on The Poisson Process. It is always interesting when a random variable from one parametric family can be transformed into a variable from another family. Both results follows from the previous result above since \( f(x, y) = g(x) h(y) \) is the probability density function of \( (X, Y) \). The last result means that if \(X\) and \(Y\) are independent variables, and \(X\) has the Poisson distribution with parameter \(a \gt 0\) while \(Y\) has the Poisson distribution with parameter \(b \gt 0\), then \(X + Y\) has the Poisson distribution with parameter \(a + b\). With \(n = 5\), run the simulation 1000 times and note the agreement between the empirical density function and the true probability density function. This is one of the older transformation technique which is very similar to Box-cox transformation but does not require the values to be strictly positive. It su ces to show that a V = m+AZ with Z as in the statement of the theorem, and suitably chosen m and A, has the same distribution as U. -2- AnextremelycommonuseofthistransformistoexpressF X(x),theCDFof X,intermsofthe CDFofZ,F Z(x).SincetheCDFofZ issocommonitgetsitsownGreeksymbol: (x) F X(x) = P(X . \(\left|X\right|\) has distribution function \(G\) given by\(G(y) = 2 F(y) - 1\) for \(y \in [0, \infty)\). Suppose that \(Z\) has the standard normal distribution, and that \(\mu \in (-\infty, \infty)\) and \(\sigma \in (0, \infty)\). we can . Suppose that \(\bs X\) is a random variable taking values in \(S \subseteq \R^n\), and that \(\bs X\) has a continuous distribution with probability density function \(f\). We will explore the one-dimensional case first, where the concepts and formulas are simplest. Using your calculator, simulate 5 values from the Pareto distribution with shape parameter \(a = 2\). Note that he minimum on the right is independent of \(T_i\) and by the result above, has an exponential distribution with parameter \(\sum_{j \ne i} r_j\). I have tried the following code: As with convolution, determining the domain of integration is often the most challenging step.