GO Classes GATE DA | Probability | Weekly Quiz 5 | Question: 9

Answer 1

Take N = 2,

So, $Var(X_{1} - X_{2} + X_{3} - X_{4}) = Var(X_{1} - X_{2}) + Var(X_{3} - X_{4}) + Cov(X_{1}-X_{2}, X_{3}-X_{4})$

Now $Var(X_{1} - X_{2}) = Var(X_{1}) + Var(X_{2}) - 2Cov(X_{1}, X_{2}) = 4 + 4 - 2*3 = \boxed{2}$

And $Var(X_{3} - X_{4}) = 2$ as $Var(X_{i}) = 4$, $Cov(X_{i}, X_{j}) = 3$

Also, $Cov(X_{1}-X_{2}, X_{3}-X_{4}) = Cov(X_{1}, X_{3}) - Cov(X_{1}, X_{4}) - Cov(X_{2}, X_{3}) + Cov(X_{2}, X_{4})$

                                                       $= 3 - 3 - 3 + 3 = 0$

Putting all values in $Var(X_{1} - X_{2} + X_{3} - X_{4}) = 2 + 2 + 0 = 4 = \boxed{2n} ......(n = 2)$

So, option B is the correct.

Answer 2

Let $X_{1}, X_{2}, \ldots, X_{2 n}$ be random variables such that
$$
\operatorname{Var}\left(X_{i}\right)=4 \quad \text { for all } i=1,2, \ldots, 2 n
$$
and
$$
\operatorname{Cov}\left(X_{i}, X_{j}\right)=3 \quad \text { for all } i \neq j .
$$

Define the sum:
$$
S=X_{1}-X_{2}+X_{3}-X_{4}+\cdots+X_{2 n-1}-X_{2 n}
$$

This can be written as:
$$
S=\sum_{i=1}^{2 n} a_{i} X_{i}, \quad \text { where } a_{i}=(-1)^{i+1}
$$

We wish to compute:
$$
\operatorname{Var}(S)=\operatorname{Var}\left(\sum_{i=1}^{2 n} a_{i} X_{i}\right)
$$

Using the formula for the variance of a linear combination of random variables:
$$
\operatorname{Var}(S)=\sum_{i=1}^{2 n} a_{i}^{2} \operatorname{Var}\left(X_{i}\right)+\sum_{\substack{i, j=1 \\ i \neq j}}^{2 n} a_{i} a_{j} \operatorname{Cov}\left(X_{i}, X_{j}\right)
$$

First term:
$$
\sum_{i=1}^{2 n} a_{i}^{2} \operatorname{Var}\left(X_{i}\right)=\sum_{i=1}^{2 n} 1 \cdot 4=2 n \cdot 4=8 n
$$

Second term:

Since $\operatorname{Cov}\left(X_{i}, X_{j}\right)=3$ for $i \neq j$, we have:
$$
\sum_{i \neq j} a_{i} a_{j} \operatorname{Cov}\left(X_{i}, X_{j}\right)=3 \sum_{i \neq j} a_{i} a_{j}
$$

Now compute $\sum_{i \neq j} a_{i} a_{j}$ using:
$$
\left(\sum_{i=1}^{2 n} a_{i}\right)^{2}=\sum_{i=1}^{2 n} a_{i}^{2}+\sum_{i \neq j} a_{i} a_{j}
$$

But $\sum_{i=1}^{2 n} a_{i}=0$, since there are equal numbers of +1 and -1 . So:
$$
0=\sum_{i=1}^{2 n} a_{i}^{2}+\sum_{i \neq j} a_{i} a_{j} \Rightarrow \sum_{i \neq j} a_{i} a_{j}=-\sum_{i=1}^{2 n} a_{i}^{2}=-2 n
$$

Therefore:
$$
\sum_{i \neq j} a_{i} a_{j} \operatorname{Cov}\left(X_{i}, X_{j}\right)=3 \cdot(-2 n)=-6 n
$$

Final variance:
$$
\operatorname{Var}(S)=8 n-6 n=2 n
$$

Comments

Here is an another method to solve it

Answer 3

Given 
\[
S = X_1 - X_2 + X_3 - X_4 + \cdots + X_{2n-1} - X_{2n}
\]
and 
\[
\operatorname{Var}(X_i) = 4, \quad \operatorname{Cov}(X_i,X_j) = 3 \text{ for } i \ne j,
\]
find $\operatorname{Var}(S)$.

---

We know that $Var(X_1+X_2)= Var(X_1)+Var(X_2)+2Cov(X_1,X_2)$

This can also be written as:

$Var(X_1) + Var(X_2) + Cov(X_1, X_2)+ Cov(X_2,X_1)= \sum_i Var(X_i)+ \sum_{i\ne j} Cov(X_i,X_j)$

In general the variance of a linear combination is given by,
\[
\operatorname{Var}(S)
= \sum_i a_i^2 \operatorname{Var}(X_i)
+ \sum_{i\ne j} a_i a_j \operatorname{Cov}(X_i,X_j).
\]

Given $\operatorname{Var}(X_i) = 4$, $\operatorname{Cov}(X_i,X_j) = 3$, and $a_i^2 = 1$, we have
\[
\operatorname{Var}(S) = 4\sum_i a_i^2 + 3\sum_{i\ne j} a_i a_j.
\]

---

First term:

Each $a_i^2 = 1$, and there are $2n$ terms:
\[
\sum_i a_i^2 = 2n.
\]

---

Second term:

Now we compute $\sum_{i\ne j} a_i a_j$ by combinatorial counting.  
\[ \sum_{i\ne j} a_i a_j = -a_1a_2 -a_2a_1 +a_1a_3 + a_3a_1 -a_2a_3 -a_3a_2 ... \]

If i & j are both odd or both even then $a_ia_j = +1$

If either i or j is even then $a_ia_j = -1$  

There are $n$ odd indices and $n$ even indices.

 

Case 1: Both $i,j$ odd
From n odd numbers select 2 numbers $= \binom{n}{2}$

These can be arranged together in 2 ways (e.g. 1,3 & 3,1)

Number of (odd, odd) pairs $= \binom{n}{2} \times 2 = n(n-1)$

Contribution: $+1 \times n(n-1)$

 

Case 2: Both $i,j$ even
From n even numbers select 2 numbers $= \binom{n}{2}$

These can be arranged together in 2 ways (e.g. 2,4 & 4,2)

Number of (even, even) pairs $= \binom{n}{2} \times 2 = n(n-1)$

Contribution: $+1 \times n(n-1)$

 

Case 3: One odd, one even
From n odd numbers select 1 number $= \binom{n}{1}$

From n even numbers select 1 number $= \binom{n}{1}$

These can be arranged together in 2 ways (e.g. 1,2 & 2,1)

Number of (odd, even) pairs $= \binom{n}{1} \binom{n}{1} \times 2 = 2n^2$

Here $a_i a_j = (+1)(-1) = -1$  

Contribution: $-1 \times 2n^2$

 

Therefore,
\[
\sum_{i\ne j} a_i a_j
= +n(n-1) + n(n-1) - 2n^2 = 2n(n-1) - 2n^2 = -2n
\]

 

Substituting into the variance formula:
\[
\begin{aligned}
\operatorname{Var}(S)
&= 4(\sum_i a_i^2) + 3(\sum_{i\ne j} a_i a_j)\\
&= 4(2n) + 3(-2n)\\
&= 8n - 6n\\
&= 2n.
\end{aligned}
\]

Hence,
\[
\operatorname{Var}(S) = 2n.
\]
 

Answer 4

This might be more small and easier to understand approach.