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More about discrete random variables in .NET Integration UPC-A Supplement 2 in .NET More about discrete random variables

More about discrete random variables using barcode integrated for .net control to generate, create upc-a supplement 2 image in .net applications. upc This property GTIN - 12 for .NET is known as the substitution law of conditional probability. To derive it, we need the observation { g(X,Y ) = z} {X = xi } = { g(xi ,Y ) = z} {X = xi }.

From this we see that P( g(X,Y ) = z. X = xi ) = P({ g(X,Y ) = z} {X = xi }) P({X = xi }) P({ g(xi ,Y ) = z} {X = xi }) = P({X = xi }) = P( g(xi ,Y ) = z. X = xi ).. We can make further simpli cations if X and Y are independent. In this case, P( g(xi ,Y ) = z X = xi ) = P( GS1 - 12 for .NET g(xi ,Y ) = z, X = xi ) P(X = xi ) P( g(xi ,Y ) = z)P(X = xi ) = P(X = xi ) = P( g(xi ,Y ) = z)..

Thus, when X and Y are independent, we can write P( g(xi ,Y ) = z X = xi ) = P( .NET upc a g(xi ,Y ) = z), and we say that we drop the conditioning. Example 3.

17 (signal in additive noise). A random, integer-valued signal X is transmitted over a channel subject to independent, additive, integer-valued noise Y . The received signal is Z = X +Y as shown in Figure 3.

5. To estimate X based on the received value Z, the system designer wants to use the conditional pmf pX. Z . Find the desired conditional pmf. (3.14). Z=X+Y Figure 3.5. Signal X subjected to additive noise Y . Solution. Let X and Y be independent, discrete, integer-valued random variables with pmfs pX and pY , respectively. Put Z := X +Y .

We begin by writing out the formula for the desired pmf pX. Z (i j) = P(X = i Z = j) = P(X = VS .NET UPC Symbol i, Z = j) P(Z = j). 3.4 Conditional probability = P(Z = j X = i)P(X = i) P(Z = j) P(Z = j X = i)pX (i) = . P(Z = j). (3.15). To continue the analysis, we use the substitution law followed by independence to write P(Z = j X = i) = P(X +Y = j X = i) = P(i +Y = j X = i) = P(Y = j i X = i) = P(Y = j i) = pY ( j i). (3.16).

This result ca Visual Studio .NET UPC Symbol n also be combined with the law of total probability to compute the denominator in (3.15).

Just write pZ ( j) =. P(Z = j X = i)P(X = i). pY ( j i)pX (i).. (3.17). In other words , if X and Y are independent, discrete, integer-valued random variables, the pmf of Z = X +Y is the discrete convolution of pX and pY . It now follows that pY ( j i)pX (i) pX. Z (i j) = , pY ( j k)pX (k). where in the d .NET UPC Symbol enominator we have changed the dummy index of summation to k to avoid confusion with the i in the numerator. The Poisson( ) random variable is a good model for the number of photoelectrons generated in a photodetector when the incident light intensity is .

Now suppose that an additional light source of intensity is also directed at the photodetector. Then we expect that the number of photoelectrons generated should be related to the total light intensity + . The next example illustrates the corresponding probabilistic model.

Example 3.18 (Poisson channel). If X and Y are independent Poisson random variables with respective parameters and , use the results of the preceding example to show that Z := X + Y is Poisson( + ).

Also show that as a function of i, pX. Z (i j) is a binom visual .net GS1 - 12 ial( j, /( + )) pmf. Solution.

To nd pZ ( j), we apply (3.17) as follows. Since pX (i) = 0 for i < 0 and since pY ( j i) = 0 for j < i, (3.

17) becomes pZ ( j) = =.
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