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Let $X1$ and $X2$ be independent normal random variables, each having mean $10$ and variance $\sigma{2}$. Which probability is larger:

a. $P(X1>15)$ or $P(X1+X2>25)$;

b. $P(X1>15)$ or $P(X1+X2>30)$?

c. Find x such that $P(X1+X2>x) = P(X1>15)$.

I have absolutely no idea how to start this problems or the appropriate formulas to use.

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1 Answer

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Here's what it occurred to me. For (a):

$X_1, X_2 \sim Normal(10, \sigma^2) \Rightarrow $ $\dfrac{X_1 - 10}{\sigma} \sim Normal(0,1), \dfrac{X_2 - 10}{\sigma} \sim Normal(0, 1)$.

$P(X_1 > 15) = P\left(\dfrac{X_1 - 10}{\sigma} > \dfrac{15 - 10}{\sigma}\right) = P\left(\dfrac{X_1 - 10}{\sigma} > \dfrac{5}{\sigma}\right) = 1 - \phi\left(\dfrac{5}{\sigma}\right)$ with $\phi$ being the CDF of a standard normal.

Now, we also know that since $X_1$ and $X_2$ are independent, $X_1 + X_2 \sim Normal(10 + 10, \sigma^2 + \sigma^2) = Normal(20, 2\sigma^2)$, so:

$P(X_1 + X_2 > 25) = P\left(\dfrac{(X_1 + X_2) - 20}{\sqrt{2}\sigma} > \dfrac{25 - 20}{\sqrt{2}\sigma}\right) = 1 - \phi\left(\dfrac{5}{\sqrt{2}\sigma}\right)$

Now we just need to compare the two results. Since $\phi$ is a monotonic increasing function (because it's a CDF), this means that $x \leq y \Rightarrow \phi(x) \leq \phi(y)$. Then since $\sigma > 0$:

$\sigma < \sqrt{2}\sigma \Rightarrow \dfrac{5}{\sqrt{2}\sigma} < \dfrac{5}{\sigma} \Rightarrow \phi\left(\dfrac{5}{\sqrt{2}\sigma}\right) < \phi\left(\dfrac{5}{\sigma}\right) \Rightarrow 1 - \phi\left(\dfrac{5}{\sigma}\right) < 1 - \phi\left(\dfrac{5}{\sqrt{2}\sigma}\right)$.

$\therefore P(X_1 > 15) < P(X_1 + X_2 > 25)$

For (b) you can do the same:

$P(X_1 + X_2 > 30) = 1 - \phi\left(\dfrac{10}{\sqrt{2}\sigma}\right)$

$\dfrac{5}{\sigma} \lessgtr \dfrac{10}{\sqrt{2}\sigma} \Rightarrow \dfrac{1}{\sigma} \lessgtr \dfrac{2}{\sqrt{2}\sigma} = \dfrac{\sqrt{2}}{\sigma} \Rightarrow \dfrac{1}{\sigma} < \dfrac{\sqrt{2}}{\sigma} \Rightarrow \dfrac{5}{\sigma} < \dfrac{10}{\sqrt{2}\sigma}$

$\Rightarrow 1 - \phi\left(\dfrac{10}{\sqrt{2}\sigma}\right) < 1 - \phi\left(\dfrac{5}{\sigma}\right)$.

$\therefore P(X_1 + X_2 > 30) < P(X_1 > 15)$

For (c) just change the value for x and make them equal.

$P(X_1 + X_2 > x) = 1 - \phi\left(\dfrac{x - 20}{\sqrt{2}\sigma}\right)$

$P(X_1 > 15) = P(X_1 + X_2 > x) \iff 1 - \phi\left(\dfrac{5}{\sigma}\right) = 1 - \phi\left(\dfrac{x - 20}{\sqrt{2}\sigma}\right) \iff \phi\left(\dfrac{5}{\sigma}\right) = \phi\left(\dfrac{x - 20}{\sqrt{2}\sigma}\right) \iff \dfrac{5}{\sigma} =\dfrac{x - 20}{\sqrt{2}\sigma} \iff x = \dfrac{5\sqrt{2}\sigma}{\sigma} + 20 = 5\sqrt{2} + 20$

$\therefore P(X_1 + X_2 > 5\sqrt{2} + 20) = P(X_1 > 15)$

Hope this helped :) (I know I wrote a lot, but I used it as an excuse to learn how to type in latex haha)

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