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i:xi B j:y j C in .NET Integration Universal Product Code version A in .NET i:xi B j:y j C

i:xi B j:y j C generate, create none none for none projects console app (2.10). pXY (xi , y j ).. (2.11). Let us now speci alize (2.11) to the case that B is the singleton set B = {xk } and C is the biggest set possible, C = IR. Then (2.

11) becomes P(X = xk ,Y IR) =. i:xi =xk j:y j IR pXY (xi , y j ).. To simplify the none for none left-hand side, we use the fact that {Y IR} := { : Y ( ) IR} = to write P(X = xk ,Y IR) = P({X = xk } ) = P(X = xk ) = pX (xk ). To simplify the double sum on the right, note that the sum over i contains only one term, the term with i = k. Also, the sum over j is unrestricted.

Putting this all together yields. pX (xk ) =. pXY (xk , y j ).. This is the same none none as (2.8) if we change k to i. Thus, the pmf of X can be recovered from the joint pmf of X and Y by summing over all values of Y .

The derivation of (2.9) is similar. Joint PMFs and independence Recall that X and Y are independent if P(X B,Y C) = P(X B) P(Y C) (2.

12). Introduction to discrete random variables for all sets B a none for none nd C. In particular, taking B = {xi } and C = {y j } shows that P(X = xi ,Y = y j ) = P(X = xi ) P(Y = y j ) or, in terms of pmfs, pXY (xi , y j ) = pX (xi ) pY (y j ). (2.

13) We now show that the converse is also true; i.e., if (2.

13) holds for all i and j, then (2.12) holds for all sets B and C. To see this, write P(X B,Y C) = = =.

i:xi B j:y j C i:xi B j:y j C i:xi B pXY (xi , y j ),. by (2.11), by (2.13),.

pX (xi ) pY (y j ),. pX (xi ). j:y j C pY (y j ). = P(X B) P(Y C). Computing probabilities with M ATLAB Example 2.15.

If X geometric0 (p) with p = 0.8, compute the probability that X takes the value of an odd integer between 5 and 13. Solution.

We must compute (1 p)[p5 + p7 + p9 + p11 + p13 ]. The straightforward solution is. p = 0.8; s = 0; none for none for k = 5:2:13 % loop from 5 to 13 by steps of 2 s = s + p k; end fprintf( The answer is %g\n ,(1-p)*s). However, we can avoid using the for loop with the commandsb p = 0.8; pvec = none for none (1-p)*p. [5:2:13]; fprintf( The answer is %g\n ,sum(pvec)).

The answer is 0. 162. In this script, the expression [5:2:13] generates the vector [5 7 9 11 13].

Next, the dot notation p. [5 7 9 11 13] means that M ATLAB should do exponentiation on each component of the vector. In this case, M ATLAB computes [p5 p7 p9 p11 p13 ].

Then each component of this vector is multiplied by the scalar 1 p. This new vector is stored in pvec. Finally, the command sum(pvec) adds up the components of the vector.

. M ATLAB programs none none are usually not compiled but run through the interpreter, loops require a lot of execution time. By using vectorized commands instead of loops, programs run much faster..

b Because 2.3 Multiple random variables Example 2.16. A light sensor uses a photodetector whose output is modeled as a Poisson( ) random variable X.

The sensor triggers an alarm if X > 15. If = 10, compute P(X > 15). Solution.

First note that. 2 15 + + . none for none 2! 15! Next, since k! = (k + 1), where is the gamma function, we can compute the required probability with the commands.
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