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Statistics

Notes Hence

 1 2 2 2
Mz(t) = exp t[ + 2 ]+ t    2    ,      .
t


1
1
2
2
But this function is the m.g.f. of N [  +  ,  +  ]. From the uniqueness property (Theorem 1
2
l 2 1 2
of Unit 10), it follows that Z has
2
N[  +  ,  +  ].
2
1 2 1 2
In the next section we shall talk about functions of more than two random variables.
5.2 Functions of More than Two Random Variables
Suppose we haven random variables X , ........, X not necessarily independent and we are
1 n
interested in finding the distribution function of a function Z = g (X , ..., X ) or the joint
1 1 1 n
distribution function of Z = g (X ,.........., X ), 1  i  r, where r is any positive integer 1  r  n. The
j i 1 n
methods described in the previous section can be extended to this general case. We will not go
into detailed description or extension of the methods. We will illustrate by a few examples.
Example 7: Suppose X , X .........X , is a random sample of size n, from a certain population.
1 2 n
We shall discuss this concept of random sampling in greater detail in Block 4 Unit 15. In the
present context it will suffice to record that the above statement is a convenient alternative way
of expressing the fact that X , X , ... X are independent and identically distributed n random
1 2 n
variables with a common distribution function F(x) which coincide with the population
distribution function (see Unit 15, Block 4). Define Z = min (X , .......,X ) and Z = max (X , ......,X ).
1 1 n n 1 n
Let us find the joint distribution of (Z , Z ).
1 n
We first note that Z  Z . Let us compute the distribution function G , z of (Z , Z ). Let (z , z ) be
1 n z i n 1 n 1 n
a fixed pair where –  < z  z < m. We first consider the case z = z . Then
1 n 1 n
G , z (z , z ) = P [Z  z , Z  Z ]
z 1 n 1 n 1 1 n n
= P[Z  z ] , since the event [Z  z ]
n n n n
implies the event [Z  z ],
1 1
= P[X  Z , 1  i  n] z and z being equal.
i n 1 n
n
=  P[X  z ], since X ’s are independent
n
i
i

i 1
n
= F (z ) .
n
Now, suppose that z < z . Then we have
1 n
G , z (z , z ) = P [Z  z , Z ,  z ]
z 1 n 1 n 1 1 n n
= P [Z  z ] – P [Z < z , Z > z ]
n n n n 1 1
= P[Z  Z ] – P[z < Z  Z  z ]
n n 1 1 n n
= P[Z  z ] – P(z < X  z for 1  i£ n)
n n 1 i n
n
n 
= P (Z  z ) – P[z  X  z ], since Xi’s are independent
n 1 i n

i 1

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