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If I have two random variables $X_1$ and $X_2$ with $X_1\sim N(520,10)$ and $X_2\sim N(500,10)$, and $X_1$, $X_2$ are both speeds of airplanes where the first one is 10 km ahead of the second one. I'm what the probability is that the second plane has not caught up to the first one after $2$ hours. I know that we can model this by saying $P(X_1-X_2\le -10)$ and the solutions suggest using the distribution of $X_1-X_2$ to be $\sim N(40,20)$. Why is the new standard deviation $20$? Before taking into account the $2$ hour passing time the standard deviation would have been $14.14$.

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With your question "Why is the new standard deviation 20" you mean the variance? The usual notation is $N(\mu,\sigma^2)$, the second entry is the variance no the standard deviation. If you sum two independent normal distributions you sum (or subtract according to the sign) the means and sum the variances…. I don't follow the question very well. With the numbers you are giving $X_1-X_2$ should distribute $N(20,20)$. Note that $-X_2\sim N(-500,10)$. – Mauricio G Tec Sep 19 '13 at 4:02
If both $X_i$ are in km/h, then you should be calculating $\Pr(2X_1-2X_2 < -10)$ for the 2-hour flight. – peterwhy Sep 19 '13 at 4:05
up vote 2 down vote accepted

The $20$ has elements of guesswork. Let $U_1$ be the distance covered by the first plane in the first hour, and $V_1$ the distance covered by the same plane in the second hour. Note that $U_1$ and $V_1$ have the same distribution as $X_1$.

Let $T_1$ be the total distance covered by the plane in $2$ hours. Then $T_1=U_1+V_1$. Define $U_2,V_2, T_2$ analogously.

If we make the rather unreasonable assumption of independence between $U_1$ and $V_1$, then the variance of the sum $U_1+V_1$ is the sum of the variances, and therefore the variance of $T_1$ is $200$.

The same argument gives $200$ as the variance of $T_2$.

The variance of $T_1-T_2$ is, assuming independence, $(1)^2\text{Var}(T_1)+(-1)^2\text{Var}(T_2)$. This is $400$, giving standard deviation $20$.

Remark: We have assumed implicitly that the variance of plane speeds was calculated by timing the plane over $1$ hour chunks. But we really don't know how variance of $100$ was arrived at. The answer above is a "justification" for the choice of $20$ for the standard deviation of the difference $T_1-T_2$ of total distances covered in $2$ hours. It has, as pointed out, a built in somewhat implausible independence assumption. The independence of $X_1$ and $X_2$ is also somewhat implausible, since both planes are subject to similar weather conditions.

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$X_i$ is given as a speed, not as a distance nor a sum of distances. – peterwhy Sep 19 '13 at 4:23
Distance covered is speed times time. But I will change the letters. – André Nicolas Sep 19 '13 at 4:28
If we break it down, we get the same answer, for the speed standard deviation was presumably average speed over a $1$ hour interval. – André Nicolas Sep 19 '13 at 5:08
Not if the standard deviaion of speed was measured using one hour runs. We could calculate the variance over say a half-hour interval by calling it $\tau^2$, and using $2\tau^2=100$. Then $4$ of these intervals ($2$ hours) have variance $\tau^2+\tau^2+\tau^2+\tau^2=200$. – André Nicolas Sep 19 '13 at 5:22

Let the std dev of the $X_i$ be given as $10$. Note that we are seeking the probability of $2X_1 - 2X_2 > -10$.

By linearity, $\Bbb E(2X_1 - 2X_2) = 2\Bbb E(X_1) - 2\Bbb E(X_2) = 40$.
Also assuming independence,
$\Bbb V(2X_1 - 2X_2) = 4 \Bbb V(X_1) + 4\Bbb V(X_2) = 800$, so $\sigma(2X_1 - 2X_2) = 20\sqrt{2}$

Hence we seek the probability of $N(40, 20\sqrt{2}) > -10$.

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@Andre Nicolas Ah! Will correct the answer. – Macavity Sep 19 '13 at 4:47
$20$ is the std dev only if we assume further that beginning of second hour the speeds are randomly selected again, independent of the earlier hour. – Macavity Sep 19 '13 at 4:57

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