linear transformation of normal distribution

Hence \[ \frac{\partial(x, y)}{\partial(u, v)} = \left[\begin{matrix} 1 & 0 \\ -v/u^2 & 1/u\end{matrix} \right] \] and so the Jacobian is $ 1/u $. Hence for $x \in \R$, $\P(X \le x) = \P\left[F^{-1}(U) \le x\right] = \P[U \le F(x)] = F(x)$. $X$ is uniformly distributed on the interval $[0, 4]$. Vary $n$ with the scroll bar and note the shape of the density function. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. The Exponential distribution is studied in more detail in the chapter on Poisson Processes. Then $ Z $ has probability density function \[ (g * h)(z) = \sum_{x = 0}^z g(x) h(z - x), \quad z \in \N \], In the continuous case, suppose that $ X $ and $ Y $ take values in $ [0, \infty) $. Vary $n$ with the scroll bar and set $k = n$ each time (this gives the maximum $V$). So the main problem is often computing the inverse images $r^{-1}\{y\}$ for $y \in T$. Then $X = F^{-1}(U)$ has distribution function $F$. As with convolution, determining the domain of integration is often the most challenging step. 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 $p$. $\P(Y \in B) = \P\left[X \in r^{-1}(B)\right]$ for $B \subseteq T$. In the discrete case, $ R $ and $ S $ are countable, so $ T $ is also countable as is $ D_z $ for each $ z \in T $. Suppose that $(X, Y)$ probability density function $f$. This is more likely if you are familiar with the process that generated the observations and you believe it to be a Gaussian process, or the distribution looks almost Gaussian, except for some distortion. Note that he minimum on the right is independent of $T_i$ and by the result above, has an exponential distribution with parameter $\sum_{j \ne i} r_j$. This is a very basic and important question, and in a superficial sense, the solution is easy. The inverse transformation is $\bs x = \bs B^{-1}(\bs y - \bs a)$. $\sgn(X)$ is uniformly distributed on $\{-1, 1\}$. The family of beta distributions and the family of Pareto distributions are studied in more detail in the chapter on Special Distributions. The formulas for the probability density functions in the increasing case and the decreasing case can be combined: If $r$ is strictly increasing or strictly decreasing on $S$ then the probability density function $g$ of $Y$ is given by \[ g(y) = f\left[ r^{-1}(y) \right] \left| \frac{d}{dy} r^{-1}(y) \right| \]. As usual, the most important special case of this result is when $ X $ and $ Y $ are independent. }, \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. Let X N ( , 2) where N ( , 2) is the Gaussian distribution with parameters and 2 . Suppose that $X$ and $Y$ are independent random variables, each having the exponential distribution with parameter 1. Let $f$ denote the probability density function of the standard uniform distribution. (2) (2) y = A x + b N ( A + b, A A T). Note that $ Z $ takes values in $ T = \{z \in \R: z = x + y \text{ for some } x \in R, y \in S\} $. $V = \max\{X_1, X_2, \ldots, X_n\}$ has probability density function $h$ given by $h(x) = n F^{n-1}(x) f(x)$ for $x \in \R$. Suppose that $X$ and $Y$ are independent random variables, each with the standard normal distribution. 116. Returning to the case of general $n$, note that $T_i \lt T_j$ for all $j \ne i$ if and only if $T_i \lt \min\left\{T_j: j \ne i\right\}$. It is also interesting when a parametric family is closed or invariant under some transformation on the variables in the family. Recall that the (standard) gamma distribution with shape parameter $n \in \N_+$ has probability density function \[ g_n(t) = e^{-t} \frac{t^{n-1}}{(n - 1)! 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 $ (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\} $. The first image below shows the graph of the distribution function of a rather complicated mixed distribution, represented in blue on the horizontal axis. $ f $ increases and then decreases, with mode $ x = \mu $. Then: X + N ( + , 2 2) Proof Let Z = X + . The distribution of $ R $ is the (standard) Rayleigh distribution, and is named for John William Strutt, Lord Rayleigh. $g(v) = \frac{1}{\sqrt{2 \pi v}} e^{-\frac{1}{2} v}$ for $ 0 \lt v \lt \infty$. However I am uncomfortable with this as it seems too rudimentary. However, when dealing with the assumptions of linear regression, you can consider transformations of . The main step is to write the event $\{Y \le y\}$ in terms of $X$, and then find the probability of this event using the probability density function of $ X $. For $y \in T$. In this particular case, the complexity is caused by the fact that $x \mapsto x^2$ is one-to-one on part of the domain $\{0\} \cup (1, 3]$ and two-to-one on the other part $[-1, 1] \setminus \{0\}$. I have to apply a non-linear transformation over the variable x, let's call k the new transformed variable, defined as: k = x ^ -2. This follows from part (a) by taking derivatives with respect to $ y $ and using the chain rule. 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. Find the probability density function of $Y$ and sketch the graph in each of the following cases: Compare the distributions in the last exercise. . Find the probability density function of the difference between the number of successes and the number of failures in $n \in \N$ Bernoulli trials with success parameter $p \in [0, 1]$, $f(k) = \binom{n}{(n+k)/2} p^{(n+k)/2} (1 - p)^{(n-k)/2}$ for $k \in \{-n, 2 - n, \ldots, n - 2, n\}$. Then run the experiment 1000 times and compare the empirical density function and the probability density function. $ f(x) \to 0 $ as $ x \to \infty $ and as $ x \to -\infty $. The PDF of $ \Theta $ is $ f(\theta) = \frac{1}{\pi} $ for $ -\frac{\pi}{2} \le \theta \le \frac{\pi}{2} $. The best way to get work done is to find a task that is enjoyable to you. For $y \in T$. Suppose that $X$ has the probability density function $f$ given by $f(x) = 3 x^2$ for $0 \le x \le 1$. Obtain the properties of normal distribution for this transformed variable, such as additivity (linear combination in the Properties section) and linearity (linear transformation in the Properties . The basic parameter of the process is the probability of success $p = \P(X_i = 1)$, so $p \in [0, 1]$. The standard normal distribution does not have a simple, closed form quantile function, so the random quantile method of simulation does not work well. Find the probability density function of $Z$. In particular, the $ n $th arrival times in the Poisson model of random points in time has the gamma distribution with parameter $ n $. $f^{*2}(z) = \begin{cases} z, & 0 \lt z \lt 1 \\ 2 - z, & 1 \lt z \lt 2 \end{cases}$, $f^{*3}(z) = \begin{cases} \frac{1}{2} z^2, & 0 \lt z \lt 1 \\ 1 - \frac{1}{2}(z - 1)^2 - \frac{1}{2}(2 - z)^2, & 1 \lt z \lt 2 \\ \frac{1}{2} (3 - z)^2, & 2 \lt z \lt 3 \end{cases}$, $ g(u) = \frac{3}{2} u^{1/2} $, for $0 \lt u \le 1$, $ h(v) = 6 v^5 $ for $ 0 \le v \le 1 $, $ k(w) = \frac{3}{w^4} $ for $ 1 \le w \lt \infty $, $g(c) = \frac{3}{4 \pi^4} c^2 (2 \pi - c)$ for $ 0 \le c \le 2 \pi$, $h(a) = \frac{3}{8 \pi^2} \sqrt{a}\left(2 \sqrt{\pi} - \sqrt{a}\right)$ for $ 0 \le a \le 4 \pi$, $k(v) = \frac{3}{\pi} \left[1 - \left(\frac{3}{4 \pi}\right)^{1/3} v^{1/3} \right]$ for $ 0 \le v \le \frac{4}{3} \pi$. In statistical terms, $ \bs X $ corresponds to sampling from the common distribution.By convention, $ Y_0 = 0 $, so naturally we take $ f^{*0} = \delta $. 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 $n$. Now we can prove that every linear transformation is a matrix transformation, and we will show how to compute the matrix. Save. Location transformations arise naturally when the physical reference point is changed (measuring time relative to 9:00 AM as opposed to 8:00 AM, for example). (iii). MULTIVARIATE NORMAL DISTRIBUTION (Part I) 1 Lecture 3 Review: Random vectors: vectors of random variables. Then $ (R, \Theta, Z) $ has probability density function $ g $ given by \[ g(r, \theta, z) = f(r \cos \theta , r \sin \theta , z) r, \quad (r, \theta, z) \in [0, \infty) \times [0, 2 \pi) \times \R \], Finally, for $ (x, y, z) \in \R^3 $, let $ (r, \theta, \phi) $ denote the standard spherical coordinates corresponding to the Cartesian coordinates $(x, y, z)$, so that $ r \in [0, \infty) $ is the radial distance, $ \theta \in [0, 2 \pi) $ is the azimuth angle, and $ \phi \in [0, \pi] $ is the polar angle. Hence \[ \frac{\partial(x, y)}{\partial(u, w)} = \left[\begin{matrix} 1 & 0 \\ w & u\end{matrix} \right] \] and so the Jacobian is $ u $. Suppose that $(X_1, X_2, \ldots, X_n)$ is a sequence of independent real-valued random variables. Assuming that we can compute $F^{-1}$, the previous exercise shows how we can simulate a distribution with distribution function $F$. Sketch the graph of $ f $, noting the important qualitative features. Random variable $T$ has the (standard) Cauchy distribution, named after Augustin Cauchy. Show how to simulate the uniform distribution on the interval $[a, b]$ with a random number. It is always interesting when a random variable from one parametric family can be transformed into a variable from another family. The general form of its probability density function is Samples of the Gaussian Distribution follow a bell-shaped curve and lies around the mean. 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 $. Suppose that $(X_1, X_2, \ldots, X_n)$ is a sequence of independent real-valued random variables, with common distribution function $F$. When $b \gt 0$ (which is often the case in applications), this transformation is known as a location-scale transformation; $a$ is the location parameter and $b$ is the scale parameter. Then $ (R, \Theta) $ has probability density function $ g $ given by \[ g(r, \theta) = f(r \cos \theta , r \sin \theta ) r, \quad (r, \theta) \in [0, \infty) \times [0, 2 \pi) \]. So $(U, V)$ is uniformly distributed on $ T $. \Only if part" Suppose U is a normal random vector. To check if the data is normally distributed I've used qqplot and qqline . and a complete solution is presented for an arbitrary probability distribution with finite fourth-order moments. Find the probability density function of $Y = X_1 + X_2$, the sum of the scores, in each of the following cases: Let $Y = X_1 + X_2$ denote the sum of the scores. Using the change of variables theorem, If $ X $ and $ Y $ have discrete distributions then $ Z = X + Y $ has a discrete distribution with probability density function $ g * h $ given by \[ (g * h)(z) = \sum_{x \in D_z} g(x) h(z - x), \quad z \in T \], If $ X $ and $ Y $ have continuous distributions then $ Z = X + Y $ has a continuous distribution with probability density function $ g * h $ given by \[ (g * h)(z) = \int_{D_z} g(x) h(z - x) \, dx, \quad z \in T \], In the discrete case, suppose $ X $ and $ Y $ take values in $ \N $. Suppose first that $F$ is a distribution function for a distribution on $\R$ (which may be discrete, continuous, or mixed), and let $F^{-1}$ denote the quantile function. The minimum and maximum transformations \[U = \min\{X_1, X_2, \ldots, X_n\}, \quad V = \max\{X_1, X_2, \ldots, X_n\} \] are very important in a number of applications. Thus suppose that $\bs X$ is a random variable taking values in $S \subseteq \R^n$ and that $\bs X$ has a continuous distribution on $S$ with probability density function $f$. $ \P\left(\left|X\right| \le y\right) = \P(-y \le X \le y) = F(y) - F(-y) $ for $ y \in [0, \infty) $. Our next discussion concerns the sign and absolute value of a real-valued random variable. With $n = 5$, run the simulation 1000 times and note the agreement between the empirical density function and the true probability density function. Suppose again that $(T_1, T_2, \ldots, T_n)$ is a sequence of independent random variables, and that $T_i$ has the exponential distribution with rate parameter $r_i \gt 0$ for each $i \in \{1, 2, \ldots, n\}$. Suppose that $Z$ has the standard normal distribution. In a normal distribution, data is symmetrically distributed with no skew. The change of temperature measurement from Fahrenheit to Celsius is a location and scale transformation. Linear transformations (or more technically affine transformations) are among the most common and important transformations. Find the probability density function of the following variables: Let $U$ denote the minimum score and $V$ the maximum score. This distribution is widely used to model random times under certain basic assumptions. $g(t) = a e^{-a t}$ for $0 \le t \lt \infty$ where $a = r_1 + r_2 + \cdots + r_n$, $H(t) = \left(1 - e^{-r_1 t}\right) \left(1 - e^{-r_2 t}\right) \cdots \left(1 - e^{-r_n t}\right)$ for $0 \le t \lt \infty$, $h(t) = n r e^{-r t} \left(1 - e^{-r t}\right)^{n-1}$ for $0 \le t \lt \infty$. A fair die is one in which the faces are equally likely. Suppose also that $X$ has a known probability density function $f$. A linear transformation of a multivariate normal random vector also has a multivariate normal distribution. In this section, we consider the bivariate normal distribution first, because explicit results can be given and because graphical interpretations are possible. This follows from part (a) by taking derivatives with respect to $ y $ and using the chain rule. Linear transformation of multivariate normal random variable is still multivariate normal. A remarkable fact is that the standard uniform distribution can be transformed into almost any other distribution on $\R$. Set $k = 1$ (this gives the minimum $U$). Beta distributions are studied in more detail in the chapter on Special Distributions. The grades are generally low, so the teacher decides to curve the grades using the transformation $ Z = 10 \sqrt{Y} = 100 \sqrt{X}$. If x_mean is the mean of my first normal distribution, then can the new mean be calculated as : k_mean = x . $X = a + U(b - a)$ where $U$ is a random number. Suppose that $X$ and $Y$ are independent and have probability density functions $g$ and $h$ respectively. f Z ( x) = 3 f Y ( x) 4 where f Z and f Y are the pdfs. In particular, it follows that a positive integer power of a distribution function is a distribution function. Suppose that $ (X, Y) $ has a continuous distribution on $ \R^2 $ with probability density function $ f $. The distribution of $ Y_n $ is the binomial distribution with parameters $n$ and $p$. Using your calculator, simulate 5 values from the exponential distribution with parameter $r = 3$. We will solve the problem in various special cases. As usual, let $ \phi $ denote the standard normal PDF, so that $ \phi(z) = \frac{1}{\sqrt{2 \pi}} e^{-z^2/2}$ for $ z \in \R $. As usual, we start with a random experiment modeled by a probability space $(\Omega, \mathscr F, \P)$. This transformation is also having the ability to make the distribution more symmetric. Part (b) means that if $X$ has the gamma distribution with shape parameter $m$ and $Y$ has the gamma distribution with shape parameter $n$, and if $X$ and $Y$ are independent, then $X + Y$ has the gamma distribution with shape parameter $m + n$. Initialy, I was thinking of applying "exponential twisting" change of measure to y (which in this case amounts to changing the mean from $\mathbf{0}$ to $\mathbf{c}$) but this requires taking . In the classical linear model, normality is usually required. 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 $f$. normal-distribution; linear-transformations. This general method is referred to, appropriately enough, as the distribution function method. = e^{-(a + b)} \frac{1}{z!} But first recall that for $ B \subseteq T $, $r^{-1}(B) = \{x \in S: r(x) \in B\}$ is the inverse image of $B$ under $r$. Order statistics are studied in detail in the chapter on Random Samples. Simple addition of random variables is perhaps the most important of all transformations. I have a pdf which is a linear transformation of the normal distribution: T = 0.5A + 0.5B Mean_A = 276 Standard Deviation_A = 6.5 Mean_B = 293 Standard Deviation_A = 6 How do I calculate the probability that T is between 281 and 291 in Python? 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 . Note that the inquality is preserved since $ r $ is increasing. Now let $Y_n$ denote the number of successes in the first $n$ trials, so that $Y_n = \sum_{i=1}^n X_i$ for $n \in \N$. The independence of $ X $ and $ Y $ corresponds to the regions $ A $ and $ B $ being disjoint. Then $U$ is the lifetime of the series system which operates if and only if each component is operating. $ G(y) = \P(Y \le y) = \P[r(X) \le y] = \P\left[X \ge r^{-1}(y)\right] = 1 - F\left[r^{-1}(y)\right] $ for $ y \in T $. As we remember from calculus, the absolute value of the Jacobian is $ r^2 \sin \phi $. Our goal is to find the distribution of $Z = X + Y$. a^{x} b^{z - x} \\ & = e^{-(a+b)} \frac{1}{z!} \, ds = e^{-t} \frac{t^n}{n!} (These are the density functions in the previous exercise). The computations are straightforward using the product rule for derivatives, but the results are a bit of a mess. $X$ is uniformly distributed on the interval $[-2, 2]$. This follows from the previous theorem, since $ F(-y) = 1 - F(y) $ for $ y \gt 0 $ by symmetry. In the usual terminology of reliability theory, $X_i = 0$ means failure on trial $i$, while $X_i = 1$ means success on trial $i$. Note that the PDF $ g $ of $ \bs Y $ is constant on $ T $. Suppose that $\bs X$ has the continuous uniform distribution on $S \subseteq \R^n$. Then, with the aid of matrix notation, we discuss the general multivariate distribution. 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. For the next exercise, recall that the floor and ceiling functions on $\R$ are defined by \[ \lfloor x \rfloor = \max\{n \in \Z: n \le x\}, \; \lceil x \rceil = \min\{n \in \Z: n \ge x\}, \quad x \in \R\]. Find the probability density function of. The linear transformation of a normally distributed random variable is still a normally distributed random variable: . The Irwin-Hall distributions are studied in more detail in the chapter on Special Distributions. Let A be the m n matrix Random component - The distribution of $Y$ is Poisson with mean $\lambda$. As before, determining this set $ D_z $ is often the most challenging step in finding the probability density function of $Z$. 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 $a + b$. Suppose that $T$ has the gamma distribution with shape parameter $n \in \N_+$. This follows directly from the general result on linear transformations in (10). By definition, $ f(0) = 1 - p $ and $ f(1) = p $. In the continuous case, $ R $ and $ S $ are typically intervals, so $ T $ is also an interval as is $ D_z $ for $ z \in T $. Show how to simulate, with a random number, the exponential distribution with rate parameter $r$. the linear transformation matrix A = 1 2 The sample mean can be written as and the sample variance can be written as If we use the above proposition (independence between a linear transformation and a quadratic form), verifying the independence of and boils down to verifying that which can be easily checked by directly performing the multiplication of and . \sum_{x=0}^z \frac{z!}{x! $g(y) = \frac{1}{8 \sqrt{y}}, \quad 0 \lt y \lt 16$, $g(y) = \frac{1}{4 \sqrt{y}}, \quad 0 \lt y \lt 4$, $g(y) = \begin{cases} \frac{1}{4 \sqrt{y}}, & 0 \lt y \lt 1 \\ \frac{1}{8 \sqrt{y}}, & 1 \lt y \lt 9 \end{cases}$. It follows that the probability density function $ \delta $ of 0 (given by $ \delta(0) = 1 $) is the identity with respect to convolution (at least for discrete PDFs). 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. 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. However, it is a well-known property of the normal distribution that linear transformations of normal random vectors are normal random vectors. Here is my code from torch.distributions.normal import Normal from torch. More generally, it's easy to see that every positive power of a distribution function is a distribution function. Suppose that $X$ has the Pareto distribution with shape parameter $a$. Let X be a random variable with a normal distribution f ( x) with mean X and standard deviation X : Expand. Using your calculator, simulate 6 values from the standard normal distribution. How could we construct a non-integer power of a distribution function in a probabilistic way? Of course, the constant 0 is the additive identity so $ X + 0 = 0 + X = 0 $ for every random variable $ X $. This chapter describes how to transform data to normal distribution in R. Parametric methods, such as t-test and ANOVA tests, assume that the dependent (outcome) variable is approximately normally distributed for every groups to be compared. Suppose that $X$ has a continuous distribution on $\R$ with distribution function $F$ and probability density function $f$. From part (b), the product of $n$ right-tail distribution functions is a right-tail distribution function. $g(u, v, w) = \frac{1}{2}$ for $(u, v, w)$ in the rectangular region $T \subset \R^3$ with vertices $\{(0,0,0), (1,0,1), (1,1,0), (0,1,1), (2,1,1), (1,1,2), (1,2,1), (2,2,2)\}$. }, \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. Recall that the Poisson distribution with parameter $t \in (0, \infty)$ has probability density function $f$ given by \[ f_t(n) = e^{-t} \frac{t^n}{n! We have seen this derivation before. Then $ (R, \Theta, \Phi) $ has probability density function $ g $ given by \[ g(r, \theta, \phi) = f(r \sin \phi \cos \theta , r \sin \phi \sin \theta , r \cos \phi) r^2 \sin \phi, \quad (r, \theta, \phi) \in [0, \infty) \times [0, 2 \pi) \times [0, \pi] \]. Recall that a standard die is an ordinary 6-sided die, with faces labeled from 1 to 6 (usually in the form of dots). The matrix A is called the standard matrix for the linear transformation T. Example Determine the standard matrices for the Expert instructors will give you an answer in real-time If you're looking for an answer to your question, our expert instructors are here to help in real-time. In this case, $ D_z = \{0, 1, \ldots, z\} $ for $ z \in \N $. I have a normal distribution (density function f(x)) on which I only now the mean and standard deviation. In the order statistic experiment, select the uniform distribution. The normal distribution is perhaps the most important distribution in probability and mathematical statistics, primarily because of the central limit theorem, one of the fundamental theorems. We will explore the one-dimensional case first, where the concepts and formulas are simplest. Work on the task that is enjoyable to you. This follows from part (a) by taking derivatives. Multiplying by the positive constant b changes the size of the unit of measurement. In the last exercise, you can see the behavior predicted by the central limit theorem beginning to emerge. Often, such properties are what make the parametric families special in the first place. $ h(z) = \frac{3}{1250} z \left(\frac{z^2}{10\,000}\right)\left(1 - \frac{z^2}{10\,000}\right)^2 $ for $ 0 \le z \le 100 $, $\P(Y = n) = e^{-r n} \left(1 - e^{-r}\right)$ for $n \in \N$, $\P(Z = n) = e^{-r(n-1)} \left(1 - e^{-r}\right)$ for $n \in \N$, $g(x) = r e^{-r \sqrt{x}} \big/ 2 \sqrt{x}$ for $0 \lt x \lt \infty$, $h(y) = r y^{-(r+1)} $ for $ 1 \lt y \lt \infty$, $k(z) = r \exp\left(-r e^z\right) e^z$ for $z \in \R$. Hence by independence, \[H(x) = \P(V \le x) = \P(X_1 \le x) \P(X_2 \le x) \cdots \P(X_n \le x) = F_1(x) F_2(x) \cdots F_n(x), \quad x \in \R\], Note that since $ U $ as the minimum of the variables, $\{U \gt x\} = \{X_1 \gt x, X_2 \gt x, \ldots, X_n \gt x\}$. See the technical details in (1) for more advanced information. The formulas above in the discrete and continuous cases are not worth memorizing explicitly; it's usually better to just work each problem from scratch. Normal distributions are also called Gaussian distributions or bell curves because of their shape. 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. Then $Y_n = X_1 + X_2 + \cdots + X_n$ has probability density function $f^{*n} = f * f * \cdots * f $, the $n$-fold convolution power of $f$, for $n \in \N$.