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I have independent random variables $X_1$, $X_2$ such that $X_1 \sim N(1,1)$ and $X_2 \sim N(2,2)$, and I'm trying to find a constant $a$ such that $a(X_1 - X_2 + 1)^2$ has a chi-squared distribution.

I'm pretty sure that the sum of $y=(X_i + X_j)^2$ generally is a chi-squared distribution. But the process is eluding me, so could use some help.

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  • $\begingroup$ A linear combination of normal variables (and constants) is normal $\endgroup$ – MPW Feb 1 '15 at 15:55
  • $\begingroup$ Do you want to have a centered chi-squared? $\endgroup$ – Math-fun Feb 1 '15 at 16:11
  • $\begingroup$ @MPW Does that mean I have to find the joint PDF of $X_1 - X_2$? $\endgroup$ – Guest Feb 1 '15 at 16:11
  • $\begingroup$ @Mehdi Yes, looking for centered chi-squared. $\endgroup$ – Guest Feb 1 '15 at 16:12
  • $\begingroup$ Are $X_1$ and $X_2$ independent? Add information about that to your question. $\endgroup$ – drhab Feb 1 '15 at 16:24
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We have $\displaystyle Y=X_1-X_2+1\sim N(0,3)$ (assuming independence). Therefore $\displaystyle \sqrt{c} Y\sim N(0,1)$ if we choose $\displaystyle c=\frac{1}{3}$, that is $\displaystyle \frac{1}{3} (X_1-X_2+1)^2 \sim \chi_{(1)}^2$.

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  • $\begingroup$ Are you using the fact that if $Z \sim N(0,1)$, then $Z^2 \sim \chi_{(1)}^2 $? $\endgroup$ – Guest Feb 1 '15 at 16:30
  • $\begingroup$ This is what I used... $\endgroup$ – Math-fun Feb 1 '15 at 16:33

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