ability distributions of S1 and S2 are fS1 1s1 2 and fS2 1s2 2 s2 , 8 0 s2 4 2s1, 0 s1 1 in .NET Implement QR Code JIS X 0510 in .NET ability distributions of S1 and S2 are fS1 1s1 2 and fS2 1s2 2 s2 , 8 0 s2 4 2s1, 0 s1 1 ability distributions of S1 and S2 are fS1 1s1 2 and fS2 1s2 2 s2 , 8 0 s2 4 2s1, 0 s1 1QR encoder on .netusing barcode generation for .net vs 2010 control to generate, create qr code 2d barcode image in .net vs 2010 applications.where b is a constant that depends on the temperature of the gas and the mass of the particle. (a) Find the value of the constant a. (b) The kinetic energy of the particle is W mV 2 2 . Find the probability distribution of W. S5-8. Suppose that X has the probability distribution fX 1x2 1, 1 x 2.NET qr-codes decoder on .netUsing Barcode reader for .net vs 2010 Control to read, scan read, scan image in .net vs 2010 applications.Find the probability distribution of the random variable Y eX. S5-9. Prove that Equation S5-3 holds when y h(x) is a decreasing function of x. S5-10. The random variable X has the probability distribution fX 1x2 x 8, 0 x 4Barcode barcode library in .netusing vs .net crystal toprint barcode in asp.net web,windows application(a) Find the joint distribution of the area of the rectangle A S1 S2 and the random variable Y S1. (b) Find the probability distribution of the area A of the rectangle. S5-12. Suppose we have a simple electrical circuit in which Ohm s law V IR holds. We are interested in the probability distribution of the resistance R given that V and I are independent random variables with the following distributions: fV 1v2 and fI 1i2 1, 1 i 2 e v, v 0Barcode barcode library for .netgenerate, create bar code none in .net projectsFind the probability distribution of Y (X 2)2. S5-11. Consider a rectangle with sides of length S1 and S2, where S1 and S2 are independent random variables. The prob-Control qrcode size in c# qr bidimensional barcode size with visual c#Find the probability distribution of R. Control qr barcode image with .netgenerate, create qr code none for .net projectsMOMENT GENERATING FUNCTIONS (CD ONLY)VS .NET qr code iso/iec18004 encoder in visual basicuse .net framework qr-code writer toembed qr-code for vb.netSuppose that X is a random variable with mean . Throughout this book we have used the idea of the expected value of the random variable X, and in fact E(X) . Now suppose that we are interested in the expected value of a particular function of X, say, g(X) X r. The expected value of this function, or E[g(X)] E(X r), is called the rth moment about the origin of the random variable X, which we will denote by r. De nition The rth moment about the origin of the random variable X is E1X 2Barcode barcode library with .netgenerate, create bar code none with .net projectsr r a x f 1x2, x .net Framework code-128c integrated in .netuse visual studio .net ansi/aim code 128 implementation toget code128 with .netX discrete (S5-7) X continuous .net Vs 2010 Crystal barcode generation for .netgenerate, create barcode none for .net projectsxr f 1x2 dx,Ucc Ean 128 drawer in .netusing barcode printing for .net control to generate, create ean / ucc - 13 image in .net applications. Notice that the rst moment about the origin is just the mean, that is, E1X2 . 1 , we can write the variFurthermore, since the second moment about the origin is E1X 2 2 2 ance of a random variable in terms of origin moments as follows:Embed delivery point barcode (dpbc) on .netuse .net vs 2010 postnet printing touse postnet 3 of 5 with .netE1X 2 2 Jasper barcode printing with javagenerate, create barcode none for java projects3E1X2 4 2 Microsoft Excel pdf 417 creation on microsoft excelusing barcode integrating for office excel control to generate, create pdf417 image in office excel applications.The moments of a random variable can often be determined directly from the de nition in Equation S5-7, but there is an alternative procedure that is frequently useful that makes use of a special function. De nition The moment generating function of the random variable X is the expected value of e tX and is denoted by MX (t). That is, MX 1t2 E1e 2Receive gs1128 for vbusing an asp.net form crystal toincoporate on asp.net web,windows applicationtx a e f 1x2, x Incoporate gs1 - 12 on .netusing sql server reporting service toattach upc barcodes with asp.net web,windows applicationX discrete (S5-8) X continuous Code 3 Of 9 generating in .netusing rdlc report todraw uss code 39 in asp.net web,windows applicationetx f 1x2 dx,Control ean-13 data for visual c# ean-13 supplement 2 data for visual c#The moment generating function MX (t) will exist only if the sum or integral in the above definition converges. If the moment generating function of a random variable does exist, it can be used to obtain all the origin moments of the random variable.Control code128 data on excel spreadsheetsto include ansi/aim code 128 and code 128 code set b data, size, image with excel barcode sdkLet X be a random variable with moment generating function MX (t). Then r d r MX 1t2 ` dt r t (S5-9)Control data matrix ecc200 image for .net c#using visual studio .net touse gs1 datamatrix barcode on asp.net web,windows applicationAssuming that we can differentiate inside the summation and integral signs, d r MX 1t2 dt r r tx a x e f 1x2, x X discrete X continuous x re tx f 1x2 dx,Now if we set t 0 in this expression, we nd that d rMX 1t2 ` dt r t E1X r 2 EXAMPLE S5-5Suppose that X has a binomial distribution, that is f 1x2 n a b px 11 x p2 n x, x 0, 1, p , n Determine the moment generating function and use it to verify that the mean and variance of the binomial random variable are np and 2 np(1 p). From the de nition of a moment generating function, we have MX 1t2tx n x a e a x b p 11 x 0 n p2 n n t x a a x b 1pe 2 11 x 0 p2 n 5-10This last summation is the binomial expansion of [pet MX 1t2 dMX 1t2 dt 3pet 11 (1 p2 4 n p)]n, so Taking the rst and second derivatives, we obtain M 1t2 X and M 1t2 X If we set t d 2MX 1t2 dt 2 npet 11 p npet 2 31 p1et 12 4 nnpet 31 p1et 12 4 n0 in M 1t2 , we obtain X M 1t2 0 t X M 1t2 0 t X np 0 in M 1t2, X which is the mean of the binomial random variable X. Now if we set t np11 Therefore, the variance of the binomial random variable is np11 1np2 2 np11 EXAMPLE S5-6Find the moment generating function of the normal random variable and use it to show that the mean and variance of this random variable are and 2, respectively. The moment generating function is MX 1t2 etx 1 12 1 12 e