If $X,Y,Z$ are independent standard normal random variables, compute $P(3X+2Y<6Z-7)$.

One way is to evaluate $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{(6z-2y-7)/3}\frac{\exp(-(x^2+y^2+z^2)/2)}{2\pi\sqrt{2\pi}}dxdydz.$$ But I don't know how to calculate this.

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    $\begingroup$ Another way is to use the fact that any linear combination of independent normal random variables is a normal random variable again. $\endgroup$ Apr 21, 2012 at 16:16
  • $\begingroup$ I was thinking something along that line. So 3X+2Y~N(0,5) and 6Z-7~N(-7,6) isn't it? Then we have to calculate $$\int_{-\infty}^{\infty}\int_{\infty}^a\frac{e^{-b^2/50}}{5\sqrt{2\pi}}\frac{e^{-(a+7)^2/72}}{6\sqrt{2\pi}}dbda$$ I can't figure out this one either. $\endgroup$
    – graidym
    Apr 21, 2012 at 16:23
  • $\begingroup$ Hint: The probability is $P[3X+2Y-6Z<-7]$. $\endgroup$ Apr 21, 2012 at 16:49
  • $\begingroup$ I think I got it. 3X+2Y-6Z~N(0,-1), so $P(3X+2Y-6Z<-7)=P(A>7)$ where $A$ is the standard normal random variable. So the answer is $1-\Phi(7)$. Is it correct? $\endgroup$
    – graidym
    Apr 21, 2012 at 16:58
  • $\begingroup$ Wait, the variance can't be negative, can it? So it should be $3X+2Y-6Z~N(0,11)$? $\endgroup$
    – graidym
    Apr 21, 2012 at 16:59

1 Answer 1


If you have two independent normally distributed random variables, then the distribution of their sum is also normally distributed, where:



You can use this to solve your problem without integrals, other than one evaluation of the error function. Actually, you don't even need that, since the $7$ turns out to give you a special result for which you probably already know the result of applying the error function.

  • $\begingroup$ To be precise, you should either insert jointly after "two" or simply replace the latter with the former. $\endgroup$
    – cardinal
    Apr 21, 2012 at 16:46
  • $\begingroup$ I have updated the answer to say "independent". I realize that that is slightly stronger than "jointly", but is likely more widely understood than "jointly". You are right that I had to say something more about the relation between the two distributions to make the statement correct. $\endgroup$
    – Mark Adler
    Apr 21, 2012 at 17:10

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