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Check out the picture below.

It's from this site: http://mathworld.wolfram.com/Convolution.html

All 3 curves (red, blue, green) are normal distributions [Edit and solution to my question (thanks Hyperplane): Turns out they are not the curves of a normal distribution. I learned "Gaussian" does not necessarily mean a probability density function]. Green is the convolution of blue and red.

My question

I know the area under all normal curves is one (axiom of probability). But it seems like the red curve has the biggest area.

I know the convolution of two normals has a variance equal to the sum of the variance of $f$ and $g$... I feel this is a bad picture, but I know I'm wrong to accuse wolfram of that. Can you fix my intuiton? Is it really just all in the tails?

Sorry if this is the dumbest question ever.

enter image description here

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3 Answers 3

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Why do you assume they are all probability densities?

To me it seems like they simply chose $g(x) = \tfrac{1}{2}f(2x)$. Hence green and blue should have the same area, but not blue and red.

I manged to recreate it in desmos. It appears that they used $f(x) = e^{-(2x)^2}$ and $g(x) =\frac{1}{2}e^{-(4x)^2}$, which gives $[f*g](x) = \frac{1}{4}\sqrt{\frac{\pi}{5}}e^{-\tfrac{16}{5}x^2}$. (Or values that are at least very close to this)

enter image description here

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  • $\begingroup$ On the website it is says they are Gaussian curves... I believe that means they are legit probability density functions with areas of 1. Or perhaps they have the normal shape but are scaled as to not have an area of 1? $\endgroup$ Dec 18, 2017 at 22:19
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    $\begingroup$ @HJ_beginner Gaussian curves is anything of the form $a e^{-b(x-c)^2}$, with $b>0$. $\endgroup$
    – Hyperplane
    Dec 18, 2017 at 22:24
  • $\begingroup$ Ahh I believe you are correct then and that resolves the area not being 1. How about the variance... wouldn't the green curve be fatter than the red curve because its variance is equal to the variance of the blue and red curves added together? For example at the points $1$ and $-1$ it seems like it should be higher than the red curve. It seems like green and red have the same standard deviation. Thanks for your help and patience. $\endgroup$ Dec 18, 2017 at 22:31
  • $\begingroup$ @Hyperplane You're right. According to the site, the red and green line are Gaussians (density functions of normal distributions), but they don't define the blue line as that. However, I am curious where you get your Gaussian curve definition from, because according to the site it is a density function of a normal distribution. $\endgroup$ Dec 18, 2017 at 22:38
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    $\begingroup$ @ThePhenotype Almost everyone calls these curves gaussian. It's only when you do probability theory when you specifically imply that the total area needs to be $1$. $\endgroup$
    – Hyperplane
    Dec 18, 2017 at 22:51
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You are right. The picture (and text) is misleading. To check that, look at the boxcar function. The area of $f$ and $g$ is definitely different

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  • $\begingroup$ Ahh, this comment was helpful. My critical mistake was thinking that the Gaussians were probability density functions. But in the example they state before, the boxcar functions are not both uniform RVs because the area under the blue box looks to be less than $1$. That clued me that this page is not focused on probabilities and then I also learned today that Gaussian does not necessarily mean a probability density curve. Thanks. $\endgroup$ Dec 18, 2017 at 22:53
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The picture is not correct as Hyperplane has shown us. So the areas are not equal.

Note that such a "huge-area-under-a-small-function-nuance" illusion does exist for other functions. We can compare the convergent integral $$\lim_{t\to\infty}\int_1^t\frac{1}{x^2}dx=\lim_{t\to\infty}(1-\frac{1}{t})=1$$ and the divergent integral $$\lim_{t\to\infty}\int_1^t\frac{1}{x}dx=\lim_{t\to\infty}\ln(t)=\infty$$

Both functions $\frac{1}{x^2}$ and $\frac{1}{x}$ are very close to $0$ for $x>4$, but still the second integral keeps adding an unbounded amount of area as $t$ gets larger.

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    $\begingroup$ There is definitely not a lot of area when the line is close to $0$, as it is exponentially decreasing. Your integrals are misleading as they picture polynomially decreasing functions. $\endgroup$
    – Hyperplane
    Dec 18, 2017 at 22:43
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    $\begingroup$ @ThePhenotype did you read my answer? I pretty much reversed engineered the parameters they used. Not any one of these curves is a probability distribution. They all have area different then 1. $\endgroup$
    – Hyperplane
    Dec 18, 2017 at 22:55
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    $\begingroup$ @HJ_beginner They even explicitely state $f$ and $g$ under the picture as density functions for normal distributions! Did any of you even read the page? $\endgroup$ Dec 18, 2017 at 23:00
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    $\begingroup$ Very, very fair point. I see I am in the wrong now. $\endgroup$ Dec 18, 2017 at 23:24
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    $\begingroup$ @ThePhenotype I still stand by my point though that even if they were all Normal distribution, there cannot be much area hiding in the tail. You should be familiar with the 68-95-99 rule? As soon are you are a few standard deviations away from the the maximum, there is almost no area left under the graph. $\endgroup$
    – Hyperplane
    Dec 18, 2017 at 23:35

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