# Do 2 Normal Distributions always produce another Normal Distribution?

This is something I am struggling to understand. I saw other proofs about this, but I want to try and do this myself using convolutions.

I always thought that the sums of two independent Gaussian Random Variables are always Gaussian. I tried to prove this using convolutions:

For two random variables $$X$$ and $$Y$$, define a new random variable $$Z = X+Y$$. Then, the distribution of this new random variable $$Z$$ can be given by the convolution :

$$r(z) = (p * q)(z) = \int_{-\infty}^{\infty} p(x) q(z - x) dx$$

If $$X$$ and $$Y$$ are both independent Gaussians, I heard that their sum is also Gaussian. I tried to prove this myself:

$$(f_X*f_Y)(z) = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi\sigma_1^2}} e^{ -\frac{(x-\mu_1)^2}{2\sigma_1^2} } \frac{1}{\sqrt{2\pi\sigma_2^2}} e^{ -\frac{(z-x-\mu_2)^2}{2\sigma_2^2} } dx$$

$$= \frac{1}{2\pi\sigma_1\sigma_2} \int_{-\infty}^{\infty} e^{ -\frac{(x-\mu_1)^2}{2\sigma_1^2} -\frac{(z-x-\mu_2)^2}{2\sigma_2^2} } dx$$

$$= \frac{1}{2\pi\sigma_1\sigma_2} \int_{-\infty}^{\infty} e^{ -\frac{1}{2} \left( \frac{(x-\mu_1)^2}{\sigma_1^2} + \frac{(z-x-\mu_2)^2}{\sigma_2^2} \right) } dx$$

From here the steps/manipulation get a bit confusing for me, but I tried to continue:

$$= \frac{1}{2\pi\sigma_1\sigma_2} \int_{-\infty}^{\infty} e^{ -\frac{1}{2} \left( \frac{x^2}{\sigma_1^2} - 2\frac{x\mu_1}{\sigma_1^2} + \frac{\mu_1^2}{\sigma_1^2} + \frac{z^2}{\sigma_2^2} - 2\frac{zx}{\sigma_2^2} + \frac{x^2}{\sigma_2^2} + \frac{\mu_2^2}{\sigma_2^2} \right) } dx$$

$$= \frac{1}{2\pi\sigma_1\sigma_2} \int_{-\infty}^{\infty} e^{ -\frac{1}{2} \left( \frac{x^2}{\sigma_1^2} + \frac{x^2}{\sigma_2^2} - 2\frac{x\mu_1}{\sigma_1^2} + \frac{\mu_1^2}{\sigma_1^2} + \frac{z^2}{\sigma_2^2} - 2\frac{zx}{\sigma_2^2} + \frac{\mu_2^2}{\sigma_2^2} \right) } dx$$

$$= \frac{1}{2\pi\sigma_1\sigma_2} \int_{-\infty}^{\infty} e^{ -\frac{1}{2} \left( x^2\left(\frac{1}{\sigma_1^2} + \frac{1}{\sigma_2^2}\right) - 2x\left(\frac{\mu_1}{\sigma_1^2} + \frac{z}{\sigma_2^2}\right) + \frac{\mu_1^2}{\sigma_1^2} + \frac{z^2}{\sigma_2^2} + \frac{\mu_2^2}{\sigma_2^2} \right) } dx$$

I think we can "complete the square". The term inside the exponent can be rewritten as:

$$-\frac{1}{2} \left( x^2\left(\frac{1}{\sigma_1^2} + \frac{1}{\sigma_2^2}\right) - 2x\left(\frac{\mu_1}{\sigma_1^2} + \frac{z}{\sigma_2^2}\right) + \frac{\mu_1^2}{\sigma_1^2} + \frac{z^2}{\sigma_2^2} + \frac{\mu_2^2}{\sigma_2^2} \right)$$

This is a quadratic expression in the form $$ax^2 + bx + c$$. To complete the square, we want to rewrite this in the form $$a(x-h)^2 + k$$. By choosing the following values we can complete the square:

$$a = \frac{1}{\sigma_1^2} + \frac{1}{\sigma_2^2}$$ $$b = -\left(\frac{\mu_1}{\sigma_1^2} + \frac{z}{\sigma_2^2}\right)$$. $$h = \frac{\mu_1/\sigma_1^2 + z/\sigma_2^2}{1/\sigma_1^2 + 1/\sigma_2^2}$$.

If we substitute these back into the previous equation, we get:

$$a(x-h)^2 = \left(\frac{1}{\sigma_1^2} + \frac{1}{\sigma_2^2}\right)\left(x - \frac{\mu_1/\sigma_1^2 + z/\sigma_2^2}{1/\sigma_1^2 + 1/\sigma_2^2}\right)^2$$

The remaining terms in the exponent can be grouped together to form $$k = \frac{\mu_1^2}{\sigma_1^2} + \frac{z^2}{\sigma_2^2} + \frac{\mu_2^2}{\sigma_2^2} - h^2$$.

Thus in total:

$$a(x-h)^2 + k = \left(\frac{1}{\sigma_1^2} + \frac{1}{\sigma_2^2}\right)\left(x - \frac{\mu_1/\sigma_1^2 + z/\sigma_2^2}{1/\sigma_1^2 + 1/\sigma_2^2}\right)^ + \frac{\mu_1^2}{\sigma_1^2} + \frac{z^2}{\sigma_2^2} + \frac{\mu_2^2}{\sigma_2^2} - h^2$$

Now, if we revisit the original convolution formula and make these substitutions:

$$(f_X*f_Y)(z) = \frac{1}{2\pi\sigma_1\sigma_2} \int_{-\infty}^{\infty} e^{ -\frac{1}{2} \left( a(x-h)^2 + k \right) } dx$$

$$(f_X*f_Y)(z) = \frac{1}{2\pi\sigma_1\sigma_2} e^{-\frac{1}{2} \left( \frac{\mu_1^2}{\sigma_1^2} + \frac{z^2}{\sigma_2^2} + \frac{\mu_2^2}{\sigma_2^2} - \left(\frac{\mu_1/\sigma_1^2 + z/\sigma_2^2}{1/\sigma_1^2 + 1/\sigma_2^2}\right)^2 \right)}$$

Using the formula $$ax^2 + bx + c$$, the term inside the exponent can be further simplified

$$-\frac{1}{2} \left( \frac{x^2}{\sigma_1^2} + \frac{x^2}{\sigma_2^2} - 2\frac{x\mu_1}{\sigma_1^2} + \frac{\mu_1^2}{\sigma_1^2} + \frac{z^2}{\sigma_2^2} - 2\frac{zx}{\sigma_2^2} + \frac{\mu_2^2}{\sigma_2^2} \right)$$

$$-\frac{1}{2} \left( x^2\left(\frac{1}{\sigma_1^2} \frac{1}{\sigma_2^2}\right) - 2x \left(\frac{\mu_1}{\sigma_1^2} + \frac{z}{\sigma_2^2}\right) + \frac{\mu_1^2}{\sigma_1^2} + \frac{z^2}{\sigma_2^2} + \frac{\mu_2^2}{\sigma_2^2} \right)$$

Here, $$a = \frac{1}{\sigma_1^2} + \frac{1}{\sigma_2^2}$$, $$b = -\left(\frac{\mu_1}{\sigma_1^2} + \frac{z}{\sigma_2^2}\right)$$, and $$c = \frac{\mu_1^2}{\sigma_1^2} + \frac{z^2}{\sigma_2^2} + \frac{\mu_2^2}{\sigma_2^2}$$.

I think I am really lost now, I am not sure how to justify that the above can be used to show that the resulting convolution is Gaussian with $$\mu = \mu_1 + \mu_2$$ and $$\sigma^2 = \sigma^2_1 + \sigma^2_2$$.

Can someone please help me out with this and continue the proof? Ideally, I would like to look at the exponent term and manipulate it until it becomes of the form: $$\frac{x-\mu}{\sigma}$$, i.e.

$$\frac{z - (\mu_1 + \mu_2)}{\sigma^2_1 + \sigma^2_2}$$

Then I could right away see that: $$\mu = \mu_1 + \mu_2$$ and $$\sigma^2 = \sigma^2_1 + \sigma^2_2$$.

Thank you!

• You can prove it by convolution and completing the square as you are, but the proof can get a little messy. A cleaner proof uses the characteristic function (the transform of convolution is a product). Commented May 8 at 19:08
• Michael: Thank you so much ... if you have time, could you please write this cleaner version of the proof using characteristic functions? I would really appreciate it! :) Commented May 8 at 19:09
• Just look in Wikipedia for the formula for the characteristic function of a Gaussian $N(\mu,\sigma^2)$. Then multiply two of those (with parameters $\mu_1,\sigma_1$ and $\mu_2,\sigma_2$), to see how the parameters add. Commented May 8 at 19:25
• You don't have enough terms in your expansion of $(z-x-\mu_2)^2$. Commented May 8 at 19:27