Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
4  
Another way is to use the fact that any linear combination of independent normal random variables is a normal random variable again. –  Raskolnikov Apr 21 '12 at 16:16
    
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. –  graidym Apr 21 '12 at 16:23
    
Hint: The probability is $P[3X+2Y-6Z<-7]$. –  David Mitra Apr 21 '12 at 16:49
    
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? –  graidym Apr 21 '12 at 16:58
    
Wait, the variance can't be negative, can it? So it should be $3X+2Y-6Z~N(0,11)$? –  graidym Apr 21 '12 at 16:59
show 1 more comment

1 Answer

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

$$\mu=\mu_1+\mu_2$$

$$\sigma^2=\sigma_1^2+\sigma_2^2$$

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.

share|improve this answer
    
To be precise, you should either insert jointly after "two" or simply replace the latter with the former. –  cardinal Apr 21 '12 at 16:46
    
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. –  Mark Adler Apr 21 '12 at 17:10
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.