# Normal rv's to chi-squared distribution

Question:

$Z_1$ and $Z_2$ are independent random variables with Normal Distributions.
$E(Z_1)$ = 1 and $V(Z_2)$ = 1; $E(Z_2)$ = 2 and $V(Z_2)$ = 2. Is there a value of k such that

$k(Z_1 - Z_2 + 1)^2$ exhibits a chi-squared distribution?

My thoughts:

Knowing that $Z_1 ~ N(1,1)$ and that $Z_2 ~ N(2,2)$ we can let $Y = k(Z_1 - Z_2 + 1)^2$. Then use the and MGF to determine the value of k: $M_Y = E(e^tY) = E[\exp{t(k(Z_1 - Z_2 +1)^2}]$

Not sure where to go from here. Am I doing this the right way? Is it much easier than this?

• Can you work out the distribution of $Z_1-Z_2+1$? Commented Sep 21, 2016 at 7:20
• Z1 and Z2 are both normal. You're just adding two normals and a constant, which should still be a normal distribution. Now, you try to square your new normal which makes a chi-square, so any positive k should do. You can't set k to negative because chi-square is non-negative. Commented Sep 21, 2016 at 7:39
• en.wikipedia.org/wiki/… what happens if you add two normal. Commented Sep 21, 2016 at 7:39
• @StudentT you didn't leave much for the OP to do there on their homework Commented Sep 21, 2016 at 10:32
• @Glen_b In this kind of question, am I supposed to solve it like how I did or just give hints? Commented Sep 21, 2016 at 10:32

I shouldn't do your homework... The key to this question is:

• What distribution you get if you add two normal random variables?
• What distribution you get if you add a constant to a normal variable?
• What distribution you get when you square a normal RV?
• What's the range of the chi-square distribution?
• +1--good hints. But could you explain why the last bullet might be relevant? It seems superfluous to me. Of greater relevance would be to ask what effect rescaling a Normal variate has on its mean and variance. (The point is that $k(Z_1-Z_2+1)^2 = (\sqrt{k}(Z_1-Z_2+1))^2$ for positive $k$.)
– whuber
Commented Sep 21, 2016 at 14:25
• @whuber The last point is important to prevent negative k. We know k can't be negative because the chi-square can't go under zero. Commented Sep 21, 2016 at 14:43
• There's no need to "prevent" negative $k$, because that will never be a solution. The question only asks whether there exists some $k$. Ruling out some subset of such values doesn't seem to help answer that in this case. It might even be distracting, because the evident intent of the question is for the student to calculate what $k$ should be and then check whether any solution is possible.
– whuber
Commented Sep 21, 2016 at 15:54