This question is about lemma 3.1.1 in Amir Dembo's note, which states as
Let $Y_{n,k}$ be mutually independent random variables, each having the normal distribution $\mathcal{N}(\mu_{n,k},\ \sigma_{n,k})$. Then $G_{n}:=\sum_{k=1}^{n}Y_{n,k}$ has the normal distribution $\mathcal{N}(\mu_{n}, \sigma_{n})$ where $\mu_{n}:=\sum_{k=1}^{n}\mu_{n,k}$ and $\sigma_{n}:=\sum_{k=1}^{n}\sigma_{n,k}$.
I am reading his proof but I got lost in the middle.
Recall that the normal distribution with mean $\mu\in\mathbb{R}$ and variance $\sigma>0$, denoted $\mathcal{N}(\mu,\sigma)$, has density function $$f(y)=\dfrac{1}{\sqrt{2\pi\sigma}}\exp\Big(-\dfrac{(y-\mu)^{2}}{2\sigma}\Big).$$
Firstly, it suffices to show the case where $n=2$, since then an inductive proof can conclude the case of $n>2$.
Also, we can WLOG suppose $Y_{n,k}\sim\mathcal{N}(0,\sigma_{n,k})$. Indeed, recall that $Y\sim\mathcal{N}(\mu,\sigma)$ if and only if $Y-\mu\sim\mathcal{N}(0,\sigma)$, and thus if $Y_{n,k}$ does not admit $\mathcal{N}(0,\sigma_{n,k})$, we can then argue with $X_{n,k}:=Y_{n,k}-\mu_{n,k}\sim\mathcal{N}(0,\sigma_{n,k})$ and to show $F_{n,k}=\sum_{k=1}^{n}X_{n,k}=G_{n}-\mu_{n}\sim\mathcal{N}(0,\sigma_{n})$, which produce the same result.
Thus, since $n=2$ is fixed, let us omit the superscript $n$, and set $G=Y_{1}+Y_{2}$, with $Y_{i}\sim\mathcal{N}(0,\sigma_{i})$. Then, we want to show that $$f_{G}(z)=\dfrac{1}{\sqrt{2\pi(\sigma_{1}+\sigma_{2})}}\exp\Big(-\dfrac{z^{2}}{2(\sigma_{1}+\sigma_{2})}\Big).$$
Recall that the density function of $G=Y_{1}+Y_{2}$ is the convolution of the density function of $Y_{1}$ with $Y_{2}$. That is $$f_{G}(z)=\int_{-\infty}^{\infty}f_{Y_{1}}(z-y)f_{Y_{2}}(y)dy=\int_{-\infty}^{\infty}\dfrac{1}{\sqrt{2\pi\sigma_{1}}}\exp\Big(\dfrac{-(z-y)^{2}}{2\sigma_{1}}\Big)\dfrac{1}{\sqrt{2\pi\sigma_{2}}}\exp\Big(\dfrac{-y^{2}}{2\sigma_{2}}\Big)dy.$$
Then, his argument becomes a little bit confusing:
Comparing this with the formula of probability density function of normal distribution for $\sigma=\sigma_{1}+\sigma_{2}$, it just remains to show that for any $z\in\mathbb{R}$, we have $$(*)\ 1=\int_{-\infty}^{\infty}\dfrac{1}{\sqrt{2\pi u}}\exp\Big(\dfrac{z^{2}}{2\sigma}-\dfrac{(z-y)^{2}}{2\sigma_{1}}-\dfrac{y^{2}}{2\sigma_{2}}\Big)dy,$$ where $u=\sigma_{1}\sigma_{2}/(\sigma_{1}+\sigma_{2})$. It is not hard to check that the argument of the exponential function is $-(y-cz)^{2}/(2u)$ for $c=\sigma_{2}/(\sigma_{1}+\sigma_{2})$. Consequently, the equation $(*)$ above is merely the obvious fact that the $\mathcal{N}(cz,u)$ density function integrates to $1$ (as any density function should), no matter what the value of $z$ is.
I am really confused here:
(1) why is it sufficient to show $(*)$?
(2) Even if I showed $(*)$, the fact that it is a density function of $\mathcal{N}(cz,u)$ integrating to $1$ seems not helping me get the density function of $G$.
What did I miss?
Thank you!
Edit 1:
I figured this out following the suggestion of "Spencer", but I don't want to give a proof that is too beautiful and formal since this kind of proof is lack of thinking process and cannot let your learn. Copying this kind of proof is to "manipulate" your proof to get close to the answer after you know the answer, which is not how we solve a problem. Therefore, I am posting my proof here:
It suffices to show the case where $n=2$, since then an inductive proof can conclude the case of $n>2$.
Also, we can WLOG suppose $Y_{n,k}\sim\mathcal{N}(0,\sigma_{n,k})$. Indeed, recall that $Y\sim\mathcal{N}(\mu,\sigma)$ if and only if $Y-\mu\sim\mathcal{N}(0,\sigma)$, and thus if $Y_{n,k}$ does not admit $\mathcal{N}(0,\sigma_{n,k})$, we can then argue with $X_{n,k}:=Y_{n,k}-\mu_{n,k}\sim\mathcal{N}(0,\sigma_{n,k})$ and to show $F_{n,k}=\sum_{k=1}^{n}X_{n,k}=G_{n}-\mu_{n}\sim\mathcal{N}(0,\sigma_{n})$, which produce the same result.
Thus, since $n=2$ is fixed, let us omit the superscript $n$, and set $G=Y_{1}+Y_{2}$, with $Y_{i}\sim\mathcal{N}(0,\sigma_{i})$. Then, we want to show that $$f_{G}(z)=\dfrac{1}{\sqrt{2\pi\sigma}}\exp\Big(-\dfrac{z^{2}}{2\sigma}\Big),$$ where $\sigma=\sigma_{1}+\sigma_{2}$.
Recall that the density function of $G=Y_{1}+Y_{2}$ is the convolution of the density function of $Y_{1}$ with $Y_{2}$. That is $$f_{G}(z)=\int_{-\infty}^{\infty}f_{Y_{1}}(z-y)f_{Y_{2}}(y)dy=\int_{-\infty}^{\infty}\dfrac{1}{\sqrt{2\pi\sigma_{1}}}\exp\Big(\dfrac{-(z-y)^{2}}{2\sigma_{1}}\Big)\dfrac{1}{\sqrt{2\pi\sigma_{2}}}\exp\Big(\dfrac{-y^{2}}{2\sigma_{2}}\Big)dy.$$
Now the proof get stuck since we don't know what to prove, but remember you want to show $$f_{G}(z)=\dfrac{1}{\sqrt{2\pi\sigma}}\exp\Big(-\dfrac{z^{2}}{2\sigma}\Big),$$ so let us identify this with the integration above to see what we need to do.
$$\int_{-\infty}^{\infty}\dfrac{1}{\sqrt{2\pi\sigma_{1}}}\exp\Big(\dfrac{-(z-y)^{2}}{2\sigma_{1}}\Big)\dfrac{1}{\sqrt{2\pi\sigma_{2}}}\exp\Big(\dfrac{-y^{2}}{2\sigma_{2}}\Big)dy=\dfrac{1}{\sqrt{2\pi\sigma}}\exp\Big(-\dfrac{z^{2}}{2\sigma}\Big),$$ since we are integrating with respect to $y$, then the RHS can be divided into the LHS, so that we need to show $$\int_{-\infty}^{\infty}\sqrt{\dfrac{\sigma_{1}+\sigma_{2}}{2\pi(\sigma_{1}\sigma_{2})}}\exp\Big(\dfrac{z^{2}}{2(\sigma_{1}+\sigma_{2})}-\dfrac{(z-y)^{2}}{2\sigma_{1}}-\dfrac{y^{2}}{2\sigma_{2}}\Big)dy=1.$$
To show this, firstly look at radical that $$\sqrt{\dfrac{\sigma_{1}+\sigma_{2}}{2\pi(\sigma_{1}\sigma_{2})}}=\dfrac{1}{\sqrt{2\pi v}},$$ where $v=(\sigma_{1}\sigma_{2})/(\sigma_{1}+\sigma_{2}),$ so an idea here is that we want to show that the things in the exponential is in the form of $-(y-\mu)^{2}/2v$ for some $\mu$ not depending on $y$, so that what inside the whole integral is a density function of $\mathcal{N}(\mu,v)$ so that the integral of it over $\mathbb{R}$ is $1$.
With a really long computation, you would have $$\dfrac{z^{2}}{2(\sigma_{1}+\sigma_{2})}-\dfrac{(z-y)^{2}}{2\sigma_{1}}-\dfrac{y^{2}}{2\sigma_{2}}=\dfrac{-(y-cz)^{2}}{2v},$$ where $c=\sigma_{2}/(\sigma_{1}+\sigma_{2})$.
Thus, the thing inside the integral is a density function of $\mathcal{N}(cz,v)$ and thus the integral is $1$. The result follows immediately.
The result you are looking for is,
$$f_G(z) = \frac{1}{\sqrt{2\pi \sigma}} \exp\Big(-\frac{z^2}{2\sigma^2}\Big)$$
If you establish $(*)$ then you can rearrange the equation to show that the convlution integral equals this result. The key step is factoring out the constant $\exp(z^2/2\sigma^2)$ and dividing it to the other side. Whats left on the right hand side will be the convolution integral times a multiplicative factor.