Prove that for any $x \in (0,\infty)$, we have Prove for all $x \geq 0$, 288.9 500 277.8 277.8 480.6 516.7 444.4 516.7 444.4 305.6 500 516.7 238.9 266.7 488.9 A generalization of gamma distribution is defined by slightly modifying the form of Kobayashi's generalized gamma function (1991). stream 585.3 831.4 831.4 892.9 892.9 708.3 917.6 753.4 620.2 889.5 616.1 818.4 688.5 978.6 ASTIN Bulletin, 192-217. \begin{align*} > * T: the random variable for wait time until the k-th event (This is the random variable of interest!) /LastChar 196 distribution, λ is the mean.) Example: Canadian Automobile Insurance Claims Source: Bailey, R.A. and Simon, LeRoy J. Thus. 12 0 obj &= \frac{\lambda^{\alpha}}{\Gamma(\alpha)} \int_0^{\infty} x^{\alpha + 1} e^{-\lambda x} {\rm d}x \\ Chi-square distribution. < Notation! (iii) The prior distribution is gamma with probability density function: (100 ) 6 100 120 e f λ λ λ λ − = (iv) Month Number of Insureds Number of Claims 1 100 6 2 150 8 3 200 11 4 300 ? /ExtGState 17 0 R /ProcSet[/PDF] 733.3 733.3 733.3 702.8 794.4 641.7 611.1 733.3 794.4 330.6 519.4 763.9 580.6 977.8 &\textrm{(using Property 2 of the gamma function)} \\ Var(X) &= EX^2 - (EX)^2 \\ In particular, the arrival times in the Poisson process have gamma distributions, and the chi-square distribution is a special case of the gamma distribution. endobj /Filter/FlateDecode We show that $I^2=2\pi$. &\textrm{(using Property 2 of the gamma function)} \\ \frac{1-q^{\large{n}}}{1-q}=1-(1-p)^n$. &\textrm{(using Property 3 of the gamma function)} \\ Suppose the number of customers arriving at a store obeys a Poisson distribution with an average of $\lambda$ 4.2 Example G revisited. For a particular machine, its useful lifetime is modeled by (f t )= 0.1 e− 0.1 t, 0 ≤ t ≤ ∞ (and 0 otherwise). 0 0 0 0 0 0 580.6 916.7 855.6 672.2 733.3 794.4 794.4 855.6 794.4 855.6 0 0 794.4 For $x \in(0,\infty)$, we have, We can write $X=\sigma Z$, where $Z \sim N(0,1)$. \end{align*} (1960). &= \frac{\alpha}{\lambda^2}. &\textrm{(using Property 3 of the gamma function)} \\ Let $X$ be the arrival $$ /Type/Font &= \frac{\alpha\Gamma(\alpha)}{\lambda\Gamma(\alpha)} /BaseFont/CMFTVE+CMSY7 1135.1 818.9 764.4 823.1 769.8 769.8 769.8 769.8 769.8 708.3 708.3 523.8 523.8 523.8 {ӏ�H0��%�����j�wY^!���c~w?�����|�Z ��z�{K�o�/-ZL��. Show that the constant in the normal distribution must be $\frac{1}{\sqrt{2 \pi}}$. &= \frac{\lambda^{\alpha}}{\Gamma(\alpha)} \int_0^{\infty} x^2 \cdot x^{\alpha - 1} e^{-\lambda x} {\rm d}x \\ /BaseFont/WUHAUO+CMSS10 646.5 782.1 871.7 791.7 1342.7 935.6 905.8 809.2 935.9 981 702.2 647.8 717.8 719.9 EX &= \int_0^\infty x f_X(x) dx \\ Let $I=\int_{-\infty}^{\infty} e^{-\frac{x^2}{2}}dx$. The distribution with p.d.f. To find $EX$ we can write /Subtype/Type1 523.8 585.3 585.3 462.3 462.3 339.3 585.3 585.3 708.3 585.3 339.3 938.5 859.1 954.4 Let $U \sim Uniform(0,1)$ and $X=-\ln (1-U)$. $=p. 794.4 794.4 702.8 794.4 702.8 611.1 733.3 763.9 733.3 1038.9 733.3 733.3 672.2 343.1 $$\lim_{\Delta \rightarrow 0} F_X(x)=1-e^{-\lambda x}.$$, If $Y \sim Geometric(p)$ and $q=1-p$, then. &= \int_0^\infty x \cdot \frac{\lambda^{\alpha}}{\Gamma{\alpha}} x^{\alpha - 1} e^{-\lambda x} {\rm d}x \\ So the purpose of this article is to provide accurate small sample inference procedures for one-sample and two-sample problems involving gamma distributions. 9 0 obj 892.9 585.3 892.9 892.9 892.9 892.9 0 0 892.9 892.9 892.9 1138.9 585.3 585.3 892.9 Find µ~ , µ and σ. time of the first customer. << 530.6 255.6 866.7 561.1 550 561.1 561.1 372.2 421.7 404.2 561.1 500 744.4 500 500 First note that since $R_U=(0,1)$, $R_X=(0,\infty)$. Calculate the Bühlmann-Straub credibility estimate of the number of claims in Month 4. 305.6 550 550 550 550 550 550 550 550 550 550 550 305.6 305.6 366.7 855.6 519.4 519.4 We have. Description: The data give the Canadian automobile insurance experience for policy years 1956 and 1957 as of June 30, 1959. Then: 1 0 00 xe xx fx x Note: This is a very useful formula when working with the Gamma distribution. Suppose that the store opens at time $t=0$. 476.4 550 1100 550 550 550 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 That is, show that /Name/F1 This is proportional to the PDF of the Gamma(s+ ;n+ ) distribution, so the posterior distribution of must be Gamma( s+ ;n+ ). 892.9 892.9 892.9 892.9 892.9 892.9 892.9 892.9 892.9 892.9 892.9 1138.9 1138.9 892.9 /Name/F3 [It plays a vital role later in understanding another important distribution, called t-distribution later.] (A) 16.7 (B) 16.9 (C) 17.3 (D) 17.6 (E) 18.0 . This should look familiar. &= \frac{(\alpha + 1)\Gamma(\alpha + 1)}{\lambda^2 \Gamma(\alpha)} >> $=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\frac{x^2+y^2}{2}}dxdy$. $=1-(1-p)^{\lfloor \frac{\Large{x}}{\Delta} \rfloor}=1-(1-\lambda \Delta)^{\lfloor \frac{\Large{x}}{\Delta} \rfloor}$. endobj >> /FirstChar 33 &= \frac{\alpha (\alpha + 1)}{\lambda^2}. 641.7 586.1 586.1 891.7 891.7 255.6 286.1 550 550 550 550 550 733.3 488.9 565.3 794.4 $$, So, we conclude &= \frac{\lambda^{\alpha}}{\Gamma(\alpha)} \frac{\Gamma(\alpha + 1)}{\lambda^{\alpha + 1}} $\lim \limits_{x \rightarrow \infty} h(x)=0$; $h'(x)=-\frac{2}{\sqrt{2\pi}}\left( \frac{e^{-\frac{x^2}{2}}}{(x^2+1)^2}\right) < 0$, for all $x \geq 0$. /LastChar 196 777.8 500 861.1 972.2 777.8 238.9 500] /Name/Im1 $=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\frac{x^2+y^2}{2}}dxdy$, $=\int_{0}^{\infty} \int_{0}^{2\pi} e^{-\frac{r^2}{2}}r d\theta dr$, $=2 \pi \int_{0}^{\infty} r e^{-\frac{r^2}{2}} dr$. Hint: Write $I^2$ as a double integral in polar coordinates. /LastChar 196 $$, Similarly, we can find $EX^2$: $$ Definition. $=2 \pi \bigg[-e^{-\frac{r^2}{2}}\bigg]_{0}^{\infty}=2 \pi$. The derivation of the PDF of Gamma distribution is very similar to that of the exponential distribution PDF, except for one thing — it’s the wait time until the k-th event, instead of the first event. /Type/XObject In Lecture 4.1, we verified that f is a probability density function, then found various probabilities. $=\lim_{\Delta \rightarrow 0} 1-(1-\lambda \Delta)^{\lfloor \frac{\Large{x}}{\Delta} \rfloor}$, $=1-\lim_{\Delta \rightarrow 0} (1-\lambda \Delta)^{\lfloor \frac{\Large{x}}{\Delta} \rfloor}$, $=\Phi\left(\frac{1-(-1)}{4}\right)-\Phi\left(\frac{(-2)-(-1)}{4}\right)$, $=\frac{1-\Phi(\frac{2-2}{2})}{1-\Phi(\frac{1-2}{2})}$, $=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} |t| e^{-\frac{t^2}{2}}dt$, $=\frac{2}{\sqrt{2\pi}}\int_{0}^{\infty} |t| e^{-\frac{t^2}{2}}dt \hspace{20pt}(\textrm{integral of an even function})$, $=\sqrt{\frac{2}{\pi}}\int_{0}^{\infty} t e^{-\frac{t^2}{2}}dt$, $=\sqrt{\frac{2}{\pi}}\bigg[-e^{-\frac{t^2}{2}} \bigg]_{0}^{\infty}=\sqrt{\frac{2}{\pi}}$, $= \int_{-\infty}^{\infty} e^{-\frac{x^2}{2}}dx \int_{-\infty}^{\infty} e^{-\frac{y^2}{2}}dy$. \end{align*} \begin{align*} 16 0 obj That is, if $Y$ is the number of customers arriving in an interval of length $t$, /BBox[0 0 2384 3370] then $Y \sim Poisson (\lambda t)$. Show that $X \sim Exponential(1)$. EX^2 &= \int_0^{\infty} x^2 {\rm d}x \\ /FirstChar 33 book homework problems are about recognizing the gamma probability density function, setting up f(x), and recognizing the mean and vari-ance ˙2 (which can be computed from and r), and seeing the connection of the gamma to the exponential and the Poisson process. /Subtype/Form endobj \end{align*} &= \frac{\lambda^{\alpha}}{\Gamma(\alpha)} \int_0^{\infty} x \cdot x^{\alpha - 1} e^{-\lambda x} {\rm d}x \\ /Subtype/Type1 However, this is one of the most common definitions of the density. Here . \begin{align*} 1138.9 1138.9 892.9 329.4 1138.9 769.8 769.8 1015.9 1015.9 0 0 646.8 646.8 769.8 Find $P(X > 1)$: We have $\mu_X=2$ and $\sigma_X=2$. Show that $X \sim Exponential(\lambda)$. 238.9 794.4 516.7 500 516.7 516.7 341.7 383.3 361.1 516.7 461.1 683.3 461.1 461.1 f(x| , ) is called Gamma distribution with parameters and and it is denoted as ( , ). Let $X \sim Gamma(\alpha,\lambda)$, where $\alpha, \lambda \gt 0$. As the prior and posterior are both Gamma distributions, the Gamma distribution is a conjugate prior for in the Poisson model. then f(x| , ) will be a probability density function since it is nonnegative and it integrates to one. &= \frac{\lambda^{\alpha}}{\Gamma(\alpha)} \frac{\Gamma(\alpha + 2)}{\lambda^{\alpha + 2}} /BaseFont/CDBYVL+CMSSBX10 Example: The Gamma distribution Suppose X has a Gamma distribution with parameters and . $$\frac{1}{\sqrt{2\pi}} \frac{x}{x^2+1} e^{-\frac{x^2}{2}} \leq P(Z \geq x) \leq \frac{1}{\sqrt{2\pi}} \frac{1}{x} e^{-\frac{x^2}{2}}.$$. /Matrix[1 0 0 1 -225 -370] A continuous random variable X is said to have a gamma distribution with parameters α > 0 and λ > 0, shown as X ∼ Gamma(α, λ), if its PDF is given by fX(x) = {λαxα − 1e − λx Γ (α) x > 0 0 otherwise If we let α = 1, we obtain fX(x) = {λe − λx x > 0 0 otherwise Thus, we conclude Gamma(1, λ) = Exponential(λ). /FontDescriptor 11 0 R The Gamma Distribution In this section we will study a family of distributions that has special importance in probability statistics.

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