1. ## Question about mean and variance

Question : If $\displaystyle X_1$ has mean 4 and variance 9 while $\displaystyle X_2$ has mean -2 and variance 5, and the two are independent, determine
i) $\displaystyle E(2X_1 + X_2 -5)$
ii) $\displaystyle Var(2X_1 + X_2 -5)$

i) $\displaystyle E(2X_1 + X_2 -5)$ = $\displaystyle 2[E(X_1)] + [E(X_2) -5]$ =$\displaystyle 8 + [-2 -5]$ =$\displaystyle 1$

ii) $\displaystyle Var(2X_1 + X_2 -5)$ = $\displaystyle 2[Var(X_1)] + [Var(X_2) -5]$ = $\displaystyle 2(9) + [5 -5]$ = $\displaystyle 18 + 0$ = $\displaystyle 18$

Is my answer correct and is my method for solving the answer crorect or no

2. Scalars a and b acting on a random variable X as:

$\displaystyle aX+b$

and act on a random variables variance as:

$\displaystyle a^2\sigma^2_x$;

Why? Think about it this way. If you had the ages of 10 people, and you multiplied their ages by 2 and added 10 years, the 10 years isn't going to affect the variation because they have all been adjusted evenly by 10 years. What will affect their ages is multiplying by the 2, and the variance is adjust by the square of the scalar.

3. Originally Posted by zorro
Question : If $\displaystyle X_1$ has mean 4 and variance 9 while $\displaystyle X_2$ has mean -2 and variance 5, and the two are independent, determine
i) $\displaystyle E(2X_1 + X_2 -5)$
ii) $\displaystyle Var(2X_1 + X_2 -5)$

i) $\displaystyle E(2X_1 + X_2 -5)$ = $\displaystyle 2[E(X_1)] + [E(X_2) -5]$ =$\displaystyle 8 + [-2 -5]$ =$\displaystyle 1$

ii) $\displaystyle Var(2X_1 + X_2 -5)$ = $\displaystyle 2[Var(X_1)] + [Var(X_2) -5]$ = $\displaystyle 2(9) + [5 -5]$ = $\displaystyle 18 + 0$ = $\displaystyle 18$

Is my answer correct and is my method for solving the answer crorect or no
$\displaystyle E(aX_1 + bX_2) = a E(X_1) + b E(X_2)$ and $\displaystyle E(cX + d) = c E(X) + d$ and so your answer to (i) is correct.

However, if $\displaystyle X_1$ and $\displaystyle X_2$ are independent then $\displaystyle Var(aX_1 + bX_2) = a^2 Var(X_1) + b^2 Var(X_2)$ and $\displaystyle Var(cX + d) = c^2 Var(X)$. Therefore your answer to (ii) is wrong.

4. ## Is this correct?

Originally Posted by mr fantastic
$\displaystyle E(aX_1 + bX_2) = a E(X_1) + b E(X_2)$ and $\displaystyle E(cX + d) = c E(X) + d$ and so your answer to (i) is correct.

However, if $\displaystyle X_1$ and $\displaystyle X_2$ are independent then $\displaystyle Var(aX_1 + bX_2) = a^2 Var(X_1) + b^2 Var(X_2)$ and $\displaystyle Var(cX + d) = c^2 Var(X)$. Therefore your answer to (ii) is wrong.

Mr fantastic is this correct

$\displaystyle Var(2X_1 + X_2 - 5)$ = $\displaystyle 2^2 [Var(X_1)] + Var(X_2) - 5$ = $\displaystyle 4 * 9 + 5 -5$ = $\displaystyle 36$ .......Is this correct

5. Originally Posted by zorro
Mr fantastic is this correct

$\displaystyle Var(2X_1 + X_2 - 5)$ = $\displaystyle 2^2 [Var(X_1)] + Var(X_2) - 5$ = $\displaystyle 4 * 9 + 5 -5$ = $\displaystyle 36$ .......Is this correct
Not quite.

$\displaystyle Var(X_2 - 5) = Var(X_2)$ (using $\displaystyle Var(cX + d) = c^2 Var(X)$ and noting that c = 1 and d = -5). Therefore $\displaystyle Var(2 X_1 + X_2 - 5) = 2^2 Var(X_1) + Var(X_2 - 5) = 2^2 Var(X_1) + Var(X_2) = ....$.

6. ## Is this correct?

Originally Posted by mr fantastic
Not quite.

$\displaystyle Var(X_2 - 5) = Var(X_2)$ (using $\displaystyle Var(cX + d) = c^2 Var(X)$ and noting that c = 1 and d = -5). Therefore $\displaystyle Var(2 X_1 + X_2 - 5) = 2^2 Var(X_1) + Var(X_2 - 5) = 2^2 Var(X_1) + Var(X_2) = ....$.

7. Originally Posted by zorro