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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

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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.

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  • $\begingroup$ 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

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