Q:

The PDF of the sum of two independent random variables with the normal distribution

I have to calculate the PDF of the sum of two independent random variables with the normal distribution. I have managed to do this using the convolution formula and calculating corresponding integral. My question is, why exactly can we set the expectation values of both PDF´s to $0$ w.l.o.g (that was a hint in the exercise)?

A:

This is an instance of a proof technique that I'll call "the Special Case implies the General Case". The assumptions are simplified for the special case, which makes it easier (more painless, less fussy) to prove. Then you use a simple device to easily extend what you've proved to the general case. (Indeed, every assertion of 'w.l.o.g.' is an appeal to this technique.)
In the context of your problem, you take advantage of the following lemma:

Lemma: If $X\sim N(\mu,\sigma^2)$ and $c$ is a constant, then $X+c\sim N(\mu+c,\sigma^2)$.

So if you've proved the Special Case:

Claim: If $X\sim N(0,\sigma_1^2)$ and $Y\sim N(0,\sigma_2^2)$ are independent, then $X+Y\sim N(0,\sigma_1^2+\sigma_2^2)$.

the lemma enables you to get for free the General Case:

Claim: If $X\sim N(\mu_1,\sigma_1^2)$ and $Y\sim N(\mu_2,\sigma_2^2)$ are independent, then $X+Y\sim N(\mu_1+\mu_2,\sigma_1^2+\sigma_2^2)$.

Proof: $X-\mu_1\sim N(0,\sigma_1^2)$ by the Lemma, and $Y-\mu_2\sim N(0,\sigma_2^2)$ by the Lemma. So by the Special Case, $$(X-\mu_1)+(Y-\mu_2)\sim N(0,\sigma_1^2+\sigma_2^2).$$ Now apply the Lemma again:
$$X+Y = (X-\mu_1) + (Y-\mu_2) + (\mu_1+\mu_2)\sim N(\mu_1+\mu_2, \sigma_1^2+\sigma_2^2),$$
and we've proved the General Case.

