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.

  • $\begingroup$ 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

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 $\displaystyle \frac{1}{3} (X_1-X_2+1)^2 \sim \chi_{(1)}^2$.

  • $\begingroup$ 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

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.