gamma distribution example problems pdf
Category : Uncategorized
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 time of the first customer. /Subtype/Type1 /Matrix[1 0 0 1 -225 -370] $$, Similarly, we can find $EX^2$: That is, if $Y$ is the number of customers arriving in an interval of length $t$, /LastChar 196 666.7 666.7 638.9 722.2 597.2 569.4 666.7 708.3 277.8 472.2 694.4 541.7 875 708.3 892.9 585.3 892.9 892.9 892.9 892.9 0 0 892.9 892.9 892.9 1138.9 585.3 585.3 892.9 $$ 855.6 550 947.2 1069.5 855.6 255.6 550] < Notation! The Gamma Distribution In this section we will study a family of distributions that has special importance in probability statistics. 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 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. 500 500 500 500 500 500 500 500 500 500 500 277.8 277.8 319.4 777.8 472.2 472.2 666.7 Let $X$ be the arrival As the prior and posterior are both Gamma distributions, the Gamma distribution is a conjugate prior for in the Poisson model. /Widths[319.4 500 833.3 500 833.3 758.3 277.8 388.9 388.9 500 777.8 277.8 333.3 277.8 f(x| , ) is called Gamma distribution with parameters and and it is denoted as ( , ). > * T: the random variable for wait time until the k-th event (This is the random variable of interest!) 16 0 obj 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. A generalization of gamma distribution is defined by slightly modifying the form of Kobayashi's generalized gamma function (1991). /Widths[1138.9 585.3 585.3 1138.9 1138.9 1138.9 892.9 1138.9 1138.9 708.3 708.3 1138.9 Example: Canadian Automobile Insurance Claims Source: Bailey, R.A. and Simon, LeRoy J. However, this is one of the most common definitions of the density. &= \frac{\alpha (\alpha + 1)}{\lambda^2}. We will find the CDF of $X$. Find $P(-2 < Y < 1)$: Since $Y=3-2X$, using Theorem 4.3, we have $Y \sim N(-1,16)$. Therefore. /FontDescriptor 11 0 R $$, $=e^{-\lambda t}\frac{(\lambda t)^0}{0!}$. /Type/Font &= \int_0^\infty x \cdot \frac{\lambda^{\alpha}}{\Gamma{\alpha}} x^{\alpha - 1} e^{-\lambda x} {\rm d}x \\ Description: The data give the Canadian automobile insurance experience for policy years 1956 and 1957 as of June 30, 1959. Var(X) &= EX^2 - (EX)^2 \\ 12 0 obj $\lim \limits_{x \rightarrow \infty} h(x)=0$; $h'(x)=-\frac{2}{\sqrt{2\pi}}\left( \frac{e^{-\frac{x^2}{2}}}{(x^2+1)^2}\right) < 0$, for all $x \geq 0$. 523.8 585.3 585.3 462.3 462.3 339.3 585.3 585.3 708.3 585.3 339.3 938.5 859.1 954.4 distribution is the Gamma distribution, i.e. Gamma Distribution Family of pdf’s that yields a wide variety of skewed distributions Distribution relies on gamma function Γ( )= Z1 0 x −1exp(−x)dx for >0 { For >1, Γ( )=( − 1)Γ( − 1) { For positive integer n,Γ(n)=(n− 1)! endobj /BaseFont/CDBYVL+CMSSBX10 $$\lim_{\Delta \rightarrow 0} F_X(x)=1-e^{-\lambda x}.$$, If $Y \sim Geometric(p)$ and $q=1-p$, then. /Name/Im1 /Subtype/Type1 /FontDescriptor 14 0 R >> &= \int_0^{\infty} x^2 \cdot \frac{\lambda^{\alpha}}{\Gamma(\alpha)} x^{\alpha - 1} e^{-\lambda x} {\rm d}x \\ EX &= \int_0^\infty x f_X(x) dx \\ $$I=\int_{-\infty}^{\infty} e^{-\frac{x^2}{2}}dx=\sqrt{2 \pi}.$$ Next, let us recall some properties of gamma function ( ). $= \frac{1}{\sqrt{2 \pi}} \int_{x}^{\infty}e^{-\frac{u^2}{2}} du$, $\leq \frac{1}{\sqrt{2 \pi}} \int_{x}^{\infty} \frac{u}{x} e^{-\frac{u^2}{2}} du \hspace{10pt} (\textrm{since $u \geq x > 0$})$, $= \frac{1}{\sqrt{2 \pi}}\frac{1}{x} \left[-e^{-\frac{u^2}{2}}\right]^{\infty}_{x}$. Show that $X \sim Exponential(1)$. 493.6 769.8 769.8 892.9 892.9 523.8 523.8 523.8 708.3 892.9 892.9 892.9 892.9 0 0 &\textrm{(using Property 2 of the gamma function)} \\ $$, So, we conclude $=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\frac{x^2+y^2}{2}}dxdy$. &= \frac{\lambda^{\alpha}}{\Gamma(\alpha)} \frac{\Gamma(\alpha + 1)}{\lambda^{\alpha + 1}} /Subtype/Form &\textrm{(using Property 3 of the gamma function)} \\ \begin{align*} 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 Calculate the Bühlmann-Straub credibility estimate of the number of claims in Month 4. Prove for all $x \geq 0$, $$\frac{1}{\sqrt{2\pi}} \frac{x}{x^2+1} e^{-\frac{x^2}{2}} \leq P(Z \geq x) \leq \frac{1}{\sqrt{2\pi}} \frac{1}{x} e^{-\frac{x^2}{2}}.$$. /FormType 1 288.9 500 277.8 277.8 480.6 516.7 444.4 516.7 444.4 305.6 500 516.7 238.9 266.7 488.9 4.2 Example G revisited. /FirstChar 33 (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 ? Two studies in automobile insurance. * Event arrivals are modeled by a Poisson process with rate λ. &= \frac{\alpha\Gamma(\alpha)}{\lambda\Gamma(\alpha)} 434.7 500 1000 500 500 500 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 << The distribution with p.d.f. (1960). Define $X=Y \Delta$, where $\lambda, \Delta>0$. /Subtype/Type1 /BBox[0 0 2384 3370] 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 $=1-(1-p)^{\lfloor \frac{\Large{x}}{\Delta} \rfloor}=1-(1-\lambda \Delta)^{\lfloor \frac{\Large{x}}{\Delta} \rfloor}$. distribution, λ is the mean.) \begin{align*} Thus, $E|X|=\sigma E|Z|$. &\textrm{(using Property 2 of the gamma function)} \\ Let $I=\int_{-\infty}^{\infty} e^{-\frac{x^2}{2}}dx$. 777.8 500 861.1 972.2 777.8 238.9 500] Thus. &\textrm{(using Property 3 of the gamma function)} \\ /LastChar 196 This should look familiar. Then: 1 0 00 xe xx fx x Note: This is a very useful formula when working with the Gamma distribution. Chi-square distribution. Find $P(X > 1)$: We have $\mu_X=2$ and $\sigma_X=2$. [It plays a vital role later in understanding another important distribution, called t-distribution later.] This is proportional to the PDF of the Gamma(s+ ;n+ ) distribution, so the posterior distribution of must be Gamma( s+ ;n+ ). 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 \end{align*} 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 \end{align*} 0 0 0 0 0 0 541.7 833.3 777.8 611.1 666.7 708.3 722.2 777.8 722.2 777.8 0 0 722.2 /FontDescriptor 8 0 R $$ $=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\frac{x^2+y^2}{2}}dxdy$, $=\int_{0}^{\infty} \int_{0}^{2\pi} e^{-\frac{r^2}{2}}r d\theta dr$, $=2 \pi \int_{0}^{\infty} r e^{-\frac{r^2}{2}} dr$. Hint: Write $I^2$ as a double integral in polar coordinates. &= \frac{\lambda^{\alpha}}{\Gamma(\alpha)} \frac{\Gamma(\alpha + 2)}{\lambda^{\alpha + 2}} Suppose that the store opens at time $t=0$. The gamma distribution f(x) = 1 2n=2( n=2) xn=2 1e x=2; x 0 with = n 2 and = 1 2 is called the chi-square distribution with ndegrees of freedom. 736.1 638.9 736.1 645.8 555.6 680.6 687.5 666.7 944.4 666.7 666.7 611.1 288.9 500 \end{align*} \frac{1-q^{\large{n}}}{1-q}=1-(1-p)^n$. Here . 238.9 794.4 516.7 500 516.7 516.7 341.7 383.3 361.1 516.7 461.1 683.3 461.1 461.1 Problem . /Name/F2 x��XI�\7��)� d��O @���H��N�o�*RSw.`x��j��L�}���U����$�o�o������p�����n�v�r�B ޕ|�]��*R\K����۟`��хKB�|�M ��\��@/��� )�*/�㔴줽@p*�ڈĘ\O/��ͻ^~�3)'��� Example: The Gamma distribution Suppose X has a Gamma distribution with parameters and . To see this, note, Let $Z \sim N(0,1)$. endobj $=\frac{1}{\sqrt{2\pi}} \frac{1}{x} e^{-\frac{x^2}{2}}$.
Spin Quantum Number, Sagemcom F@st 5260, Best Chicken Parmesan Recipe, What Are The Benefits Of Drinking Ensure?, Benefits Of Reading, Jeadv Publication Fee, 1 John 3:23-24 Esv, Baked Fish Parmigiana, Peanut Butter No Bake Cookies,
