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Let $X_1$ and $X_2$ be independent random variables with distributions, with $X_1 \sim N(-1; 2)$ and $X_2 \sim N(2; 3)$

Prove that random variable $Y_2=((2X_1+X_2)^2)/11$ is chi-square distributed.

My approach:
$2X_1+X_2 =Z,$ with $Z \sim N(-2+2;4*2+3)=N(0;11).$
$Y_2 \sim (N(0,11)^2)/11 =(N(0;1)^2)*(\sqrt{11}/11).$ Where am I making mistake?

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    What's your definition of the chi square distribution? Is it the square of $N(0,1)$? – 2017-01-05
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    Yes, I'm not sure how to handle the number outside the distribution. Is it correct way how to write? Am I supposed to transform it somehow? – 2017-01-05
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    One easy mistake that I could see would be assuming that you're writing $N(\mu,\sigma^2)$ instead of $N(\mu,\sigma)$. (In particular this would mean that you're assuming $X_1$ has variance $2$ when in fact it was intended for it to have standard deviation $2$). Check your reference for that... – 2017-01-05
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    @Ian I'm not sure I follow, might be language barrier, but in class normal distribution was explained as N(μ,σ^2), where σ^2 is the variance of distribution. If I then have N(0;11), what is the proper way to factor out that 11, because I need N(0,1) if I am not mistaken. – 2017-01-05
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    I have made an edit to your post for readability, but I had trouble making sense of the last line. Please re-edit if necessary. – 2017-01-06

1 Answers 1

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You have shown that you know the necessary facts. But I Think you may be trying to take too many steps at once. Here is a start and an outline how to finish.

I would start out by finding the distribution of $2X_1.$

$E(2X_1) = 2E(X_1) = 2(-1) = -2.$

$Var(2X_1) = 4Var(X_1) = 4(2) = 8.$

So $2X_1 \sim N(-2, 8).

Next find the distribution of $2X_1 + X_2.$

$E(2X_1 + X_2) = E(2X_1) + E(X_2) = (-2) + 2 = 0.$ Then find $Var(2X_1 + X_2),$ and hence the distribution.

Next, standardize $2X_1 + X_2$ to get $Z \sim N(0,1),$ standard normal. Finally, $Z^2$ is a chi-squared distribution (with 1 degree of freedom).

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    So the actual mistake is that they should really have written $(2X_1+X_2)^2/11= \left ( \frac{2X_1+X_2}{\sqrt{11}} \right )^2$, noting that the inside of the square is N(0,1). – 2017-01-06