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Let $X_1, X_2, X_3$ be three random variable following a normal distribution $N(6,4)$. What is the probability that the largest observation exceeds 8? Hint: $Y = \max(X_1, X_2, X_3)$.

Here is what I tried:

$P(Y\leq y) = P(X_i\leq y)$ for $i=1,2,3$. Then $$ P(Y\leq y) = P(X_1 \leq y)*P(X_2 \leq y) * P(X_3 \leq y). $$ To get $P(X_ i \leq y)$, I use the $P(X > 8) = 1 - P(X < 8) = 1 - 0.841 = 0.159$. Then I plug in $0.159$ to the previous equation: $P(X_1 \leq y) * P(X_2 \leq y) *P(X_3 \leq y)$, so that $(0.159)(0.159)(0.159) = 0.0040$.

However, my teacher said this is wrong.

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You should try to tell us what you've tried so far and where you get stuck. (In other words, we're not going to simply do your homework for you.) –  Patrick Coulombe Mar 23 at 6:38
what I did was: Pr(Y< or = y) = Pr(Xi< or = y, i=1,2,3). Then Pr(Y< or = y) = Pr(X1< or =y)xPr(X2< or =y)xPr(X3< or =y). To get the Pr(Xi< or = y), I use the Pr(x>8)=1-Pr(x<8)=1-0.841=0.159. Then I plug-in 0.159 to the previous equation: Pr(X1< or =y)xPr(X2< or =y)xPr(X3< or =y), so that (0.159)(0.159)(0.159) = 0.0040. My teacher said it it wrong. :( –  user42404 Mar 23 at 7:37
Wow, thanks for the edited version. –  user42404 Mar 23 at 8:01

2 Answers 2

up vote 4 down vote accepted

There is just a small mistake in the development provided in the question. The mistake comes from the fact that $$ P(Y > y) \neq P(X_1 > y) \cdot P(X_2 > y) \cdot P(X_3 > y) . $$ Here is a way to solve the problem. Since $P(Y > y) = 1 - P(Y \leq y)$, and assuming the mutual independence of $X_1, X_2, X_3$ \begin{align*} P(Y \leq y) &= P\left\{\max(X_1, X_2, X_3) \leq y \right\} \\ &= P(X_1 \leq y) \cdot P(X_2 \leq y) \cdot P(X_3 \leq y) \\ &= P(X_1 \leq y)^3, \end{align*} then the requested probability is $P(Y > y) = 1 - P(X_1 \leq y)^3$.

The numerical result is $P(Y > 8) = 1 - P(X_1 \leq 8)^3 = 1 - 0.8413^3 \approx 0.4044$

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I assumed that $X_i \sim N(6, 4)$ meant ${\rm var}(X_i) = 4$ and used pnorm(8, 6, 2) in R. –  QuantIbex Mar 23 at 10:19
Thanks! It took me days and still can't figure it out what to do about it. Anyways, again thanks. –  user42404 Mar 23 at 11:06

I got to the following, please let me know if there is anything wrong.

Assume that $\mu=6$ and $\sigma=4$, then we have $P(X_i \geq 8)= 0.3085$ for an arbitrary $i$, i.e. $i=1,2,3$. Thus the probability that we get a number higher than 8 from an observation drawn from a sample $X_{i}$ is $0.3085$, so the probability that there is at least one observation from our sample $X_{i}$ that is larger than $8$ is equal to $P(K\geq1)=1-P(K=0)$, where $K$ is equal to the number of observations larger than 8, thus $P(K\geq1)=1-{{3}\choose{0}}\cdot 0.6915^3$, which is about $0.66934$. Well, this holds for an arbitrary $i$.

Now we would like to know what the probability is that one of these samples have an observation higher than 8. The probability that one sample has an observation higher than 8 is equal to $0.66934$. This can be done with a binomial distribution, so $P(L\geq 1)$, where $L$ stands for the number of $X_{i}$'s that have an observation larger than 8. $P(L\geq 1)=1-P(L=0)$. I get a probability of about 0.97.

Not sure whether I have made a mistake, hopefully it is helpful.

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Thanks for your help. Actually the answer is 0.405, I just don't know how to arrive to such answer. It seems just a simple normal distribution problem to me, but the Y=max(X1, X2, X3) makes life complicated. How did you get to the P(Xi≥8)=0.3085? –  user42404 Mar 23 at 8:59

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