Finding $P(X_1+X_2 > 1.9X_3)$ where $X_1$, $X_2$, and $X_3$ are independent, normal distributed random variables

Question

Suppose that $X_1$, $X_2$, and $X_3$ are independent, normal distributed with same mean and standard deviation. How do I find $P(X_1+X_2 > 1.9X_3)$?

What I did so far:

$$P(X_1+X_2 > 1.9X_3) = 1- P\left(\dfrac{X_1+X_2}{X_3}\le1.9\right)$$

Adding $X_1$ and $X_2$ together will yield a new random variable with different normal distribution:

$$X_1+X_2\sim N(\mu_{X_1}+\mu_{X_2},\sigma^2_{X_1}+\sigma^2_{X_2})$$ Then how do I calculate a random variable with ratio distribution of $\dfrac{X_1+X_2}{X_3}$? Cauchy distribution would not work here because the means are not zero. Is there other way to do this?

Answer 1

The key observation you need to make is that sums of independent Normal variables are themselves Normal.

If we define $Z = X_1 + X_2 - 1.9 X_3$ with $X_1,X_2,X_3 \sim N(\mu, \sigma^2)$ then

$$Z \sim N((1 + 1 - 1.9) \mu, (1^2 + 1^2 + (-1.9)^2)\sigma^2),$$

i.e. $Z \sim N(0.1\mu, 5.61\sigma^2)$.

And then

$$ \mathbf P [ X_1 + X_2 > 1.9X_3] = \mathbf P[ Z > 0] = 1 - \Phi \left( \frac{-0.1\mu}{\sqrt{5.61\sigma^2}} \right)= \Phi \left( \frac{0.1\mu}{\sqrt{5.61\sigma^2}} \right), $$ where $\Phi$ is the CDF of the standard Normal Distribution.

Answer 2

$X_1+X_2-(1.9)X_3$ is normal with mean $(0.1)m$ and variance $(5.61) \sigma^{2}$ where $m$ and $\sigma ^{2}$ are the mean and variance of the $X_i$'s. Write $Y$ for $X_1+X_2-(1.9)X_3$. What you want to find is $P\{Y>0\}$. The ansqer id $1-\Phi (-\frac {(0.1)m)} {\sigma\sqrt (5.61)}$