# 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;
𝑛1
𝑛1
𝑛1
𝑖=1
𝑖=1
𝑖=1
𝑛2
𝑛2
𝑛2
𝑖=1
𝑖=1
𝑖=1
1
∑𝑛𝑖=1
𝑋1𝑖
1
1
1
̅
𝐸(𝑋1 ) = 𝐸 (
) = 𝐸 (∑ 𝑋1𝑖 ) = ∑ 𝐸(𝑋1𝑖 ) = ∑ 𝐸(𝑋1𝑖 )
𝑛1
𝑛1
𝑛1
𝑛1
𝑛1
=
1
𝑛1 𝜇1
= 𝜇1
∑ 𝜇1 =
𝑛1
𝑛1
𝑖=1
2
∑𝑛𝑖=1
𝑋2𝑖
1
1
1
𝐸(𝑋̅2 ) = 𝐸 (
) = 𝐸 (∑ 𝑋2𝑖 ) = ∑ 𝐸(𝑋2𝑖 ) = ∑ 𝐸(𝑋2𝑖 )
𝑛2
𝑛2
𝑛2
𝑛2
𝑛2
=
1
𝑛2 𝜇2
= 𝜇2
∑ 𝜇2 =
𝑛2
𝑛2
𝑖=1
𝐸(𝑋̅1 − 𝑋̅2 ) = 𝐸(𝑋̅1 ) − 𝐸(𝑋̅2 ) = 𝜇1 − 𝜇2
𝑛1
𝑛1
𝑛1 𝑛1
𝑖=1
𝑖=1
𝑖=1 𝑗=1
1
1 2
1
1
𝑉(𝑋̅1 ) = 𝑉 (∑ 𝑋1𝑖 ) = ∑ ( ) 𝑉(𝑋1𝑖 ) + 2 ∑ ∑ ( ) ( ) 𝐶𝑜𝑣(𝑋𝑖 , 𝑋𝑗 )
𝑛1
𝑛1
𝑛1 𝑛1
𝑖&lt;𝑗
𝑛1
𝑛1 𝑛1
𝑖=1
𝑖=1 𝑗=1
1 2 2
1
1
1 2
𝜎1 2
2
= ∑ ( ) 𝜎1 + 2 ∑ ∑ ( ) ( )(0) = ( ) (𝑛1 𝜎1 ) + 0 =
𝑛1
𝑛1 𝑛1
𝑛1
𝑛1
𝑖&lt;𝑗
(𝑆𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 𝑋1𝑖 𝑎𝑛𝑑 𝑋1𝑗 𝑎𝑟𝑒 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑐𝑜𝑣(𝑋1𝑖 , 𝑋1𝑗 ) = 0 )
𝑛2
𝑛2
𝑛2 𝑛2
𝑖=1
𝑖=1
𝑖=1 𝑗=1
1
1 2
1
1
𝑉(𝑋̅2 ) = 𝑉 (∑ 𝑋2𝑖 ) = ∑ ( ) 𝑉(𝑋2𝑖 ) + 2 ∑ ∑ ( ) ( ) 𝐶𝑜𝑣(𝑋𝑖 , 𝑋𝑗 )
𝑛2
𝑛2
𝑛2 𝑛2
𝑖&lt;𝑗
𝑛2
𝑛2 𝑛2
𝑖=1
𝑖=2 𝑗=1
1 2 2
1
1
1 2
𝜎2 2
2
= ∑ ( ) 𝜎1 + 2 ∑ ∑ ( ) ( )(0) = ( ) (𝑛2 𝜎2 ) + 0 =
𝑛2
𝑛2 𝑛2
𝑛2
𝑛2
𝑖&lt;𝑗
(𝑆𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 𝑋2𝑖 𝑎𝑛𝑑 𝑋2𝑗 𝑎𝑟𝑒 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑐𝑜𝑣(𝑋2𝑖 , 𝑋2𝑗 ) = 0 )
𝐸(𝑋̅1 − 𝑋̅2 ) = (1)2 𝑉(𝑋̅1 ) + (−1)2 𝑉(𝑋̅2 ) + 2(1)(−1)𝐶𝑜𝑣(𝑋̅1 , 𝑋̅2 )
= 𝑉(𝑋̅1 ) + 𝑉(𝑋̅2 ) + 2(1)(−1)(0)
𝑆𝑖𝑛𝑐𝑒 𝑋̅1 𝑑𝑒𝑝𝑒𝑛𝑑𝑠 𝑜𝑛𝑙𝑦 𝑜𝑛 𝑋1𝑖 ′𝑠𝑎𝑛𝑑 𝑋̅2 𝑑𝑒𝑝𝑒𝑛𝑑𝑠 𝑜𝑛𝑙𝑦 𝑜𝑛 𝑋2𝑖 ′𝑠
(̅
)
𝑋1 𝑎𝑛𝑑 𝑋̅2 𝑎𝑟𝑒 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑎𝑛𝑑 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡ℎ𝑒𝑖𝑟 𝑐𝑜𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒 𝑖𝑠 0
𝜎1 2 𝜎1 2
= 𝑉(𝑋̅1 ) + 𝑉(𝑋̅2 ) =
+
𝑛1
𝑛2
```