# Use Theorem 4.14 and its corollary to show that if X11, X12,...,X1n1

```Use Theorem 4.14 and its corollary to show that if X11, X12,...,X1n1,X21,X22,...,X2n2 are
independent random variables, with the first n1 constituting a random sample from an
infinite population with the mean mu1 and the variance sigma 1 squared and the other n2
constituting a random sample from an infinite population with the mean mu2 and the
variance sigma2 squared, then a)E(X1-X2) (both X's have bars on top) = mu1-mu2; b)
var(X1-X2) (both X's have bars on top) = sigma1 squared divided by n1 plus sigma2
squared divided by n2 Theorem 4.14: If X1, X2...Xn are random variables and Y=the sum of
aiXi i=1 where a1, a2, and an are constants, then E(Y)= the sum of ai E(Xi) i=1 and var(Y)
= the sum of ai squared times var(Xi)+2 sigma sigma aiaj times cov (XiXj) where the
double summation extends over all values of i and j, from 1 to n, which i&lt; Thanks! possible.
if before him used I as again Pillai Rajeevan use to like would&gt;
π1
π1
π1
π=1
π=1
π=1
π2
π2
π2
π=1
π=1
π=1
1
∑ππ=1
π1π
1
1
1
Μ
πΈ(π1 ) = πΈ (
) = πΈ (∑ π1π ) = ∑ πΈ(π1π ) = ∑ πΈ(π1π )
π1
π1
π1
π1
π1
=
1
π1 π1
= π1
∑ π1 =
π1
π1
π=1
2
∑ππ=1
π2π
1
1
1
πΈ(πΜ2 ) = πΈ (
) = πΈ (∑ π2π ) = ∑ πΈ(π2π ) = ∑ πΈ(π2π )
π2
π2
π2
π2
π2
=
1
π2 π2
= π2
∑ π2 =
π2
π2
π=1
πΈ(πΜ1 − πΜ2 ) = πΈ(πΜ1 ) − πΈ(πΜ2 ) = π1 − π2
π1
π1
π1 π1
π=1
π=1
π=1 π=1
1
1 2
1
1
π(πΜ1 ) = π (∑ π1π ) = ∑ ( ) π(π1π ) + 2 ∑ ∑ ( ) ( ) πΆππ£(ππ , ππ )
π1
π1
π1 π1
π&lt;π
π1
π1 π1
π=1
π=1 π=1
1 2 2
1
1
1 2
π1 2
2
= ∑ ( ) π1 + 2 ∑ ∑ ( ) ( )(0) = ( ) (π1 π1 ) + 0 =
π1
π1 π1
π1
π1
π&lt;π
(πππππ π‘βπ π£ππππππππ  π1π πππ π1π πππ πππππππππππ‘ πππ£(π1π , π1π ) = 0 )
π2
π2
π2 π2
π=1
π=1
π=1 π=1
1
1 2
1
1
π(πΜ2 ) = π (∑ π2π ) = ∑ ( ) π(π2π ) + 2 ∑ ∑ ( ) ( ) πΆππ£(ππ , ππ )
π2
π2
π2 π2
π&lt;π
π2
π2 π2
π=1
π=2 π=1
1 2 2
1
1
1 2
π2 2
2
= ∑ ( ) π1 + 2 ∑ ∑ ( ) ( )(0) = ( ) (π2 π2 ) + 0 =
π2
π2 π2
π2
π2
π&lt;π
(πππππ π‘βπ π£ππππππππ  π2π πππ π2π πππ πππππππππππ‘ πππ£(π2π , π2π ) = 0 )
πΈ(πΜ1 − πΜ2 ) = (1)2 π(πΜ1 ) + (−1)2 π(πΜ2 ) + 2(1)(−1)πΆππ£(πΜ1 , πΜ2 )
= π(πΜ1 ) + π(πΜ2 ) + 2(1)(−1)(0)
πππππ πΜ1 πππππππ  ππππ¦ ππ π1π ′π πππ πΜ2 πππππππ  ππππ¦ ππ π2π ′π
(Μ
)
π1 πππ πΜ2 πππ πππππππππππ‘ πππ π‘βπππππππ π‘βπππ πππ£πππππππ ππ  0
π1 2 π1 2
= π(πΜ1 ) + π(πΜ2 ) =
+
π1
π2
```