Probability with Independent Normal Variables

Let X and Y be independent standard normal variables. Find:

a) P(3X + 2Y > 5)

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I see you haven't accepted any of the answers to your previous questions -- did they not answer your questions? –  joriki Apr 14 '11 at 6:54
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1 Answer

Here are some hints that should go a long way towards answers.

(a) In your course, you may have learned that if $X$ and $Y$ are independent normal with means $p$, $q$ and variances $s^2$, $t^2$, then $aX+bY$ is normal mean $ap+bq$, variance $a^2s^2+b^2t^2$. So $3X+2Y$ has mean $0$, variance $13$.

(b) Our probability is $1$ minus the probability that they are both $\ge 1$. What is the probability that $X \ge 1$?

(c) The probability that the absolute value of $\min(X,Y)$ is less than $1$ is the probability that $(X < 1) \cup (Y<1)$ minus the probability that $(X \le -1) \cup (Y \le -1)$.

(d) This is the probability that $|X-Y|<1$. Let $W=X-Y$ and look at the remark for part (a).

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The hint on part (a) is false: standard gaussian random variables have mean $0$ and variance $1$ hence $3X+2Y$ has mean $0$ and variance $13$, and one very much needs to know the variance. The hint on part (c) is false as well: if $X=10$ and $Y=0$, $|\min(X,Y)|=0$ but $X$ does not lie between $-1$ and $1$. (And already three upvotes... :-)) –  Did Apr 15 '11 at 14:33
@Didier Pau: I thank you for pointing out the errors. There was further discussion in another post, where I corrected (a). Is there a way of removing the upvotes? –  André Nicolas Apr 19 '11 at 16:24
You are welcome. Please do not worry about upvotes (or downvotes, for that matter), they mean next to nothing. –  Did Apr 19 '11 at 18:15
But you definitely should correct your answer (why you do not is beyond me). –  Did Apr 24 '11 at 10:23
@Didier Piau: I had corrected once, completely, but it did not take. Maybe I forgot to save. I am not satisfied with my hint in (c), it is less intuitively clear than the others. –  André Nicolas Apr 24 '11 at 20:17
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