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I have two independent normal distributions with different mean and variance, such that.

$$f_1(x_1 \; | \; \mu_1, \sigma_1) = \frac{1}{\sigma_1\sqrt{2\pi} } \; \exp({ -\frac{(x-\mu_1)^2}{2\sigma_1^2} })$$

$$f_2(x_2 \; | \; \mu_2, \sigma_2) = \frac{1}{\sigma_2\sqrt{2\pi} } \; \exp({ -\frac{(x-\mu_2)^2}{2\sigma_2^2} })$$

This is an example I made in R considering:

$\mu_{1} = 1$, $\sigma^{2}_{1} = 1.4$

$\mu_{2} = 3.5$, $\sigma^{2}_{2} = 1.1$

Distribution comparison

I tried to analytically find the mathematical expression of the gray area, that is the common mass shared by both distributions. I have looked at other posts, like this one, but as far as I understand the difference between two normals is a normal distribution. However, in this plot this doesn't look like a normal distribution.

I would say something like

$f_{3}(x_1,x_2 | \mu_1, \sigma_1,\mu_2, \sigma_2) = \int(f_{1}-f_{2})$

but I am afraid this is not correct.

I'd appreciate some intuition and a mathematical expression for the gray area.

Stephan Kolassa
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adrian1121
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    This is done in two easy steps. First, you solve for the peak of the intersection. (By taking logarithms, this reduces to solving a quadratic equation: and that ought to alert you to the possibility that the gray area might actually be *multiple* gray areas in some circumstances!) Then, with that solution in hand, you may partition the region into tail regions whose areas are given by the cumulative distribution functions. – whuber May 30 '17 at 13:42
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    These are two different things. The graph shows the overlapping area of the two regions while the difference of two normal random variable is something quite different. – Michael R. Chernick May 30 '17 at 15:19
  • If you mean the overlapping area your question is a duplicate of wolfies link. – Michael R. Chernick May 30 '17 at 15:23

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