Normal Bell Curves Displaced by One Standard Deviation

Question

I was just researching some statistics and ran into this problem:

a)If we have two normal bell curves, displaced by one standard deviation, how large is the area of overlap?

b)What is the proportion of the area of overlap to the remaining area of one of the bell curves?

Thank you for any help in advance.

Answer 1

These things are better explained visually.

Notice that the overlap can be divided into two equal sections, one of which is shaded green.

Notice that the intersection of the bell curves happens at $x=0.5$, or more generally half a standard deviation above the first curve.

Therefore, the green area is equal to $P(z<-0.5)=0.3085$.

And the total overlap is $0.717$.

The proportion of the overlap to the remaining area is $\displaystyle \frac{0.717}{0.283} \approx \boxed{2.533}$

Answer 2

Argument for two non-standard Gaussians

Say the two curves are $$f_k(x) = \frac{1}{\sqrt{ 2 \pi \sigma_k^2}}exp(-\frac{1}{2\sigma_k^2}(x - \mu_k)^2) \qquad k \in \lbrace 1,2 \rbrace$$ Let us find the point of intersection $$f_1(x_0) = f_2(x_0)$$ i.e. $$\frac{1}{\sqrt{ 2 \pi \sigma_1^2}}exp(-\frac{1}{2\sigma_1^2}(x_0 - \mu_1)^2)=\frac{1}{\sqrt{ 2 \pi \sigma_2^2}}exp(-\frac{1}{2\sigma_2^2}(x_0 - \mu_2)^2)$$ Take logarithms on both sides, $$-\frac{1}{2} \ln 2\pi\sigma_1^2 -\frac{1}{2\sigma_1^2}(x_0 - \mu_1)^2=-\frac{1}{2} \ln 2\pi\sigma_2^2 -\frac{1}{2\sigma_2^2}(x_0 - \mu_2)^2 $$ Multiply by $-2$, i.e $$ \ln 2\pi\sigma_1^2 +\frac{1}{\sigma_1^2}(x_0 - \mu_1)^2= \ln 2\pi\sigma_2^2 +\frac{1}{\sigma_2^2}(x_0 - \mu_2)^2 $$ or $$ \ln \frac{\sigma_1^2}{\sigma_2^2} = [\frac{1}{\sigma_2}(x_0 - \mu_2)]^2 -[\frac{1}{\sigma_1}(x_0 - \mu_1)]^2$$ Continue the math to find an expression of the point $x_0$ that satisfies the above equation. As for the area, assume $\mu_1 < \mu_2$ without loss of generality, the area will be taken from $-\infty$ (on the tail of $f_2(x)$ till the point of intersection, then we exchange from $f_2(x)$ onto $f_1(x)$ at $x_0$ till $+\infty$, i.e. $$A = \int_{-\infty}^{x_0} f_2(x) \ dx + \int_{x_0}^{+\infty} f_1(x) \ dx$$ The proportion w.r.t to each of the Bell curves is $$P_1 = \frac{A}{\int_{-\infty}^{x_0} f_1(x) \ dx}$$ and $$P_1 = \frac{A}{\int_{x_0}^{+\infty} f_2(x) \ dx}$$