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Unformatted text preview: ST 561: Important topics from Ch. 4 Transformations of a single variable: Let X be a continuous random variable, 9(X) an
increasing or decreasing function. If g is increasing: Peon s t) = For 5 g“(t)) ears"1(a).
where F}; is the c.d.f. of X, so that the p.d.f. of Y = g(X) is given by M15 = §E(Q_1(t))fx(9_l(t)} If g is decreasing,
P(9(X) S t) = P(X 2 9—16)) = 1  Fx(y“l(t)),
so the p.d.f. of Y = g(X) is given by Mt) : §£(g'1(t))fx(9'1(t)) In either case. we have (1
13/05): 3E(9'1(t)fx(9‘1(t)) If g is not monotonic, we may still be able to ﬁnd the p.d.f. of Y = 9(X) in some cases. For example,
if 9(93) = 332.
P(X2 s t) = P(—t 5 X 5 t) = Fx(t) — Fx(«t), so that the density is fxz (t) = fx (t) + fx(t). See p. 194 for some more general situations. Transformations of more than one variable: A special case is the sum of two or more independent random variables. If X1,X2 are independent, the p.d.f. of X1 + X2 is given by the
convolution of fX1 and fxz: ‘ fX1+X2(t) = f” mama — was: = f_°° meals — adm In some cases we may also use moment generating functions to ﬁnd the distribution of a sum of
random variables: MX1+...X;¢ (t) = Hic=lMXi (i), so that if we can recognize the m.g.f. of the sum X1 + ...+ Xk, We will know its distribution.
Examples of families that are closed under summation of independent variables are the normal and
Poisson families. When X1,.. . ., X k are not independent, we may need to ﬁnd the distribution of Y1, . . . , Yk, Where
1/5. = gi(X11X21'1Xk)13= 1,2,  ' *5 ‘16 Our algorithm for this goes as follows: (1) Write down the joint distribution of X1, . . . , Xk, noting the support of this distribution.
(2) Identify the support of Y1, . . . ,Yk. (3) Find the inverse transformation In, . . . , hk mapping Y1, . . . , 1’}. back into X1, . . . ,Xk.
(4) Compute the Jacobian J of this inverse transformation. (5) The joint p.d.f. of Y1, . . . , Yk is then given by fY(yls “13%): fX(h1(y11' ‘iyk)1 '  whiten,” u'ykDIJl. where J[ will in general be a function of 3,11, . . . , yk. Sampling distributions: Let .X1,...,X1C be i.i.d. with mean p. and variance 02. The sample
mean and sample variance of this random sample are given by a. = 0/11)th 32 =(1/(n — 1)) 2(a — 5...)? i=1 i=2! Whatever the distribution of the X;, we have
E(5T:n) = a, E(sz) = 02, so that :2", and 52 are unbiased estimators of a and 0'2, respectively. When the sample is from a
N (p, 0'2) distribution, we can assert that: (l) The distribution of in is N01, 02/17.). (2) The distribution of Ln—jg is chi—squared with nl degrees of freedom.
(3) 5'" and 32 are independent. The statistic Eff—"“1 has a t—distribatioa with (nl) degrees of freedom. In general, if W, U, V
are independent random variables and W is standard normal, U is chisquare with m degrees of
freedom, V is chi—square with n degrees of freedom, then: (1) The random variable (7%; has a t—distribution with 11 degrees of freedom;
(2) The random variable 51;: has an Fdistribution with (m,n) degrees of freedom. ...
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 Fall '08
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