I've taking a test and got fail in this exercise. It says:

"Let $X_1,X_2,X_3$ and $X_4$ be independent, identically distributed random variables such taht each of them has a normal distribution with mean 0 and variance 1. We have $Y_1,Y_2$ and $Y_3$ which are independent identically distributed random variables with mean 2 and variance 3. Firther, all $X_i$ and $Y_i$ are independent. Let $\overline{X}$ and $\overline{Y}$ denote the sample means of $X_i$ and $Y_j$ respectively. Calculate $V(\overline{X}-2\overline{Y})$"

(a) 1.25

(b) 4.25

(c) -5

(d) 7

I know that $V(\overline{X}-2\overline{Y})=V(\overline{X})-2V(\overline{Y})$ and then I know $V(\overline{X})=1$ because of the information in the exercise, similar for $V(\overline{Y})=3$ hence $V(\overline{X}-2\overline{Y})=1-2\cdot 3=-5$" but it's not true.

Any help would be nice


Suppose $n$ iid random variables share a variance $v$, then the variance of their mean is not $v$ but $v/n$. It follows that $V(\overline X)=\frac14$ and $V(\overline Y)=1$. Finally, multiplying a random variable by $a$ multiplies its variance by $a^2$, and variances always add. Thus the answer is $\frac14+2^2×1=4.25$.

You should have rejected $-5$ immediately, since variances are always nonnegative.

  • $\begingroup$ Auch! Thats true we can't have a variance which is negative... I understand your calculations. $\endgroup$ – Joey Adams Oct 20 '18 at 14:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.