(x m) f (x) dx in .NET Integrate UPC-A in .NET (x m) f (x) dx

(x m) f (x) dx using barcode generator for visual .net control to generate, create upc-a supplement 5 image in visual .net applications. iPad exp 1 x m 2 = (x GTIN - 12 for .NET m) 2 . 1 x m x m exp 2 = 2 Make the change of var .net vs 2010 UPC Symbol iable y = (x m)/ , noting that dy = dx/ . Then e y /2 dy, E[X m] = y 2 .

which is seen to be ze ro once we recognize the integral as having the form of the mean of an N(0, 1) random variable Y . To compute var(X), write E[(X m)2 ] =. (x m)2 f (x) dx exp 1 x m 2 = (x m)2 2 = . x m exp 1 x m 2 2 Continuous random variables Making the same change of variable as before, we obtain E[(X m)2 ] = 2 e y /2 y2 dy. 2 Now recognize this int GS1 - 12 for .NET egral as having the form of the second moment of an N(0, 1) random variable Y . By the previous example, E[Y 2 ] = 1.

Hence, E[(X m)2 ] = 2 . Example 4.13 (in nite expectation).

Pareto densitiesb have been used to model packet delay, les sizes, and other Internet characteristics. Let X have the Pareto density f (x) = 1/x2 , x 1. Find E[X].

Solution. Write E[X] =. 1 dx = x2 1 dx = ln x x = .. Example 4.14. Determin UPC A for .

NET e E[X] if X has a Cauchy density with parameter = 1. Solution. This is a trick question.

Recall that as noted following Example 2.24, for signed discrete random variables, E[X] =. i:xi 0 xi P(X = xi ) + i:xi <0 xi P(X = xi ),. if at least one of the sums is nite. The analogous formula for continuous random variables is E[X] =. x f (x) dx + x f (x) dx,. assuming at least one .net vs 2010 UPC-A of the integrals is nite. Otherwise we say that E[X] is unde ned.

Since f is the Cauchy(1) density, x f (x) = x Since this integrand has anti-derivative 1 ln(1 + x2 ), 2 we have E[X] = 1 ln(1 + x2 ) 2 . 1/ . 1 + x2 1 ln(1 + x2 ) 2 . 0 . = ( 0) + (0 ) = = unde ned. b Additional Pareto densities are considered in Problems 2, 23, and 26 and in Problem 59 in 5. 4.2 Expectation of a single random variable Derivation of LOTUS Let X be a continuous upc barcodes for .NET random variable with density f . We rst show that if g is a real-valued function taking nitely many distinct values y j IR, then E[g(X)] = To begin, observe that P(Y = y j ) = P(g(X) = y j ) = Then write E[Y ] = = = = =.

{x:g(x)=y j } . g(x) f (x) dx. (4.6). f (x) dx. y j P(Y = y j ). {x:g(x)=y j }. f (x) dx {x:g(x)=y j }. y j f (x) dx g(x) f (x) dx j . {x:g(x)=y j }. g(x) f (x) dx,. since the last sum of integrals is just a special way of integrating over all values of x. We would like to apply (4.6) for more arbitrary functions g.

However, if g(X) is not a discrete random variable, its expectation has not yet been de ned! This raises the question of how to de ne the expectation of an arbitrary random variable. The approach is to approximate X by a sequence of discrete random variables (for which expectation was de ned in 2) and then de ne E[X] to be the limit of the expectations of the approximations. To be more precise about this, consider the sequence of functions qn sketched in Figure 4.

9.. n qn ( x ). 1 ___ 2n Figure 4.9. Finite-ste .

net vs 2010 Universal Product Code version A p quantizer qn (x) for approximating arbitrary random variables by discrete random variables. The number of steps is n2n . In the gure, n = 2 and so there are eight steps.

. Continuous random variables Since each qn takes on upc a for .NET ly nitely many distinct values, qn (X) is a discrete random variable for which E[qn (X)] is de ned. Since qn (X) X, we then de ne5 E[X] := limn E[qn (X)].

Now suppose X is a continuous random variable. Since qn (X) is a discrete random variable, (4.6) applies, and we can write E[X] := lim E[qn (X)] = lim.

n n qn (x) f (x) dx. Now bring the limit in side the integral,6 and then use the fact that qn (x) x. This yields E[X] =. n . lim qn (x) f (x) dx = x f (x) dx. The same technique can be used to show that (4.6) holds even if g takes more than nitely many values. Write E[g(X)] := lim E[qn (g(X))].

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