Intersection of two gaussian distribution functions

[hkBattousai]

Can you please give me formulas which give intersection of two gaussian function

f(x, mu, sigma) = 1/sqrt(2*pi*sigma^2) * exp(-(x-mu)^2 / (2*sigma^2))

for the case variances are different.


(Note: I think it is time I learn how to use "tex" tags, do you know a good tutorial?)

 

[mathman]

What do you mean by "intersection"? The term intersection is usually used when talking about sets, not functions.

 

[hkBattousai]

What do you mean by "intersection"?

The two points, where these two curves have the same (x, y) pairs on the graph.

Well, actually, I calculated the formula, but it is too messy without using "tex" tags...

 

[Redbelly98]

Why not just set the two Gaussian expressions equal to each other, and solve for x?

Info on [noparse][tex][/noparse] tag usage at Physics Forums can be found https://www.physicsforums.com/showthread.php?t=386951"

 

[blue_raver22]

Sure, I would be happy to provide you with the formulas for the intersection of two Gaussian distribution functions. The intersection of two Gaussian functions occurs at the point where the two curves intersect, or in other words, where the two functions have the same value. This can be found by setting the two functions equal to each other and solving for x.

Let's say we have two Gaussian functions, f(x, mu1, sigma1) and g(x, mu2, sigma2). The intersection of these two functions would be given by:

f(x, mu1, sigma1) = g(x, mu2, sigma2)

1/sqrt(2*pi*sigma1^2) * exp(-(x-mu1)^2 / (2*sigma1^2)) = 1/sqrt(2*pi*sigma2^2) * exp(-(x-mu2)^2 / (2*sigma2^2))

To solve for x, we can take the natural logarithm of both sides to eliminate the exponential terms:

ln(1/sqrt(2*pi*sigma1^2)) - (x-mu1)^2 / (2*sigma1^2) = ln(1/sqrt(2*pi*sigma2^2)) - (x-mu2)^2 / (2*sigma2^2)

Next, we can rearrange the terms to isolate x on one side:

(x-mu1)^2 / (2*sigma1^2) - (x-mu2)^2 / (2*sigma2^2) = ln(1/sqrt(2*pi*sigma2^2)) - ln(1/sqrt(2*pi*sigma1^2))

We can then combine the terms on the left side using the difference of squares formula:

((x-mu1)^2 - (x-mu2)^2) / (2*sigma1^2 * sigma2^2) = ln(1/sqrt(2*pi*sigma2^2)) - ln(1/sqrt(2*pi*sigma1^2))

Next, we can simplify the right side by using the properties of logarithms:

ln(1/sqrt(2*pi*sigma2^2)) - ln(1/sqrt(2*pi*sigma1^2)) = ln(sqrt(sigma1^2/sigma2^2))

Finally, we can solve for x by multiplying both sides by (2