## Introduction to Mathematical Statistics |

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Page 12

follows that 1 = P(A) + P(A*), which is the desired result. Theorem 2. The

probability of the null set is zero, that is, P(0) = 0.

so that ...

**Proof**. We have z = A U A* and A n A* = 0. Thus, from (c) and (b) of Definition 7, itfollows that 1 = P(A) + P(A*), which is the desired result. Theorem 2. The

probability of the null set is zero, that is, P(0) = 0.

**Proof**. In Theorem 1, take A = 0so that ...

Page 47

Prsu(x) = c s ==

continuous type; but the

integrals by sums. Let A = {x; u(z) > c) and let f(z) denote the p.d.f. of X. Then Eu(X

)] ...

Prsu(x) = c s ==

**Proof**. The**proof**is given when the random variable X is of thecontinuous type; but the

**proof**can be adapted to the discrete case if we replaceintegrals by sums. Let A = {x; u(z) > c) and let f(z) denote the p.d.f. of X. Then Eu(X

)] ...

Page 72

and X2 is fi(zi) fa(z2). Thus, we have, by definition of mathematical expectation, in

the continuous case, Eu(x)/(x)] = s.s., u(r)e(...)f(r)s.(...) dr, dr, = ||... u(x)/(x) ...

**Proof**. The stochastic independence of X1 and X2 implies that the joint p.d.f. of X1and X2 is fi(zi) fa(z2). Thus, we have, by definition of mathematical expectation, in

the continuous case, Eu(x)/(x)] = s.s., u(r)e(...)f(r)s.(...) dr, dr, = ||... u(x)/(x) ...

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