SOLVED - Question involving the normal distribution

Question

I have some variable that's normally distributed with mean A and standard deviation B (variance B^2).

I make two (independent) observations. What's the probability that these observations will deviate from each other by at least a standard deviation B?

Answer 1

The difference is normally distributed as well, with $\mu_{x_1-x_2}=\mu_{x_1}-\mu_{x_2}=\mu_y=0$ and $\sigma^2_{x_1-x_2}=\sigma_{x_1}^2+\sigma_{x_2}^2=\sigma_y^2=2B^2$.

Therefore $X_1-X_2=Y\sim \mathcal N(0, 2B^2)$. Now we calculate the probability that the deviation of the absolute value of $Y$ is more than $B$. For this purpose we can standardize Y:

$Z=\frac{Y-\mu_y}{\sigma_y}=\frac{Y-0}{\sqrt 2\cdot B}$

Since $Z$ is symmetric around $0$ we have $P(|Z|>B)=2-2\Phi(B)$.

Thus $P(|Z|>B)=2-2\Phi\left(\frac{B}{\sqrt 2\cdot B} \right)= 2-2\Phi\left(\frac{1}{\sqrt 2} \right)=2-2\Phi\left(0.707107\right)$

Using a calculator for thes (standard) normal distribution we get $\Phi\left(0.707107\right)=0.76$

$P(|Z|>B)=2-2\cdot 0.76=2-1.52=0.48$

Answer 2

Let $X_i$ be IID $N(\mu, \sigma^2)$. Then $X_1-X_2$ is $N(0, 2\sigma^2)$.

Proof: Note that, using independence, $$\mathbb{E}(e^{t(X_1-X_2)})=\mathbb{E}(e^{tX_1})\mathbb{E}(e^{-tX_2})$$ $$=e^{\mu t+0.5\sigma^2 t^2}\cdot e^{-\mu t +0.5\sigma^2 t^2}=e^{\sigma^2 t^2},$$ which is the MGF of a $N(0,2\sigma^2)$ RV.

Thus, $P(X_1-X_2>\sigma)=P(Z>\tfrac{1}{\sqrt{2}})\approx 0.23975$. By the way, linear combinations of independent normal RVs are very useful. Highly suggest googling a bit deeper or reviewing your textbook, etc.