what is the difference between mixture of two normal distributions and sum of two independent variables The following denotes a mixture of a standard normal with a
normal with the same mean but 100 times the variance:
$0.95 \mathrm{~N}(0,1)+ 0.05 \mathrm{~N}(0,100)$
Let Y = 0.95 X + 0.05 Z with X,Z are independent normal distribution with X ~ N(0,1) Z~N(0,100)
I remember Y = X + Z if X ~ N(u1,s1) and Z ~ N(u2,s2), then Y would follow N(u1 + u2,s1+s2)
These two conclusions seems different and makes me confused.
Hope someone can explain
 A: If $X_1 \sim \mathsf{Norm}(\mu_1,\sigma_1)$ and, independently,
$X_2 \sim \mathsf{Norm}(\mu_1,\sigma_2)$ then
$$X_1+X_2 \sim \mathsf{Norm}\left(\mu_1+\mu_2,\sqrt{\sigma_1^2+\sigma_2^2}\right).$$
In other words, it is the variances that add,
not the standard deviations.
Also, there are some things about mixture distributions you may not understand. Although
I try to illustrate a 50-50 mixture of two
normal distributions (with different standard
deviations) below, you might want to look at the
Wikipedia page on 'mixture distributions' and
the 'Related' links in the margin of this page
that mention 'mixture' in their titles.
[The Wikipedia page gives examples of mixture
distributions with different means, but the formulas for PDFs, means, and variances are relevant.]
Because it is easier to visualize a 50:50 mixture
graphically than a 95:5 mixture, I use a 50:50 mixture below. [In the R functions for normal
distributions, the parameters are mean and SD.]
set.seed(2022); n = 10^5
p = rbinom(n, 1, .5)     # for 50:50 mixture
x1 = rnorm(n, 0, 1)
x2 = rnorm(n, 0, 10)
y.mix = p*x1 + (1-p)*x2
y.sum = x1 + x2

For samples of size $n = 100,000,$ sample means
and standard deviations should match population means and SDs to a couple of significant digits.
summary(y.mix); var(y.mix)
     Min.   1st Qu.    Median      Mean   3rd Qu.      Max. 
-40.11645  -1.52578   0.00029   0.01102   1.53021  48.13074 
[1] 50.40555   # sample SD

summary(y.sum); var(y.sum)
     Min.   1st Qu.    Median      Mean   3rd Qu.      Max. 
-42.67832  -6.75347   0.00270  -0.00142   6.81791  48.53083 
[1] 100.6077

In the figure below the (blue) histograms nearly
approximate the (black) density curves of the
distributions for the mixture and the sum.
par(mfrow=c(2,1))
hist(y.mix, prob=T, br=100, col="skyblue2")
 curve(.5*dnorm(x,0,1)+.5*dnorm(x,0,10), add=T, lwd=3)
hist(y.sum, prob=T, br=100, col="skyblue2")
 curve(dnorm(x, 0, sqrt(1+100)), add=T, lwd=3)
par(mfrow=c(1,1))


If you have two independent normal random variables, then their sum is normal, as shown at the beginning of this Answer. However, the mixture of two independent
normal random variables need not be even nearly normal.
A: Another way to think about it: if $X$ is generated from one distribution and $Y$ from another, then $X+Y$ is a sample from the distribution of the sum.  To get a sample from the mixture, you pick either $X$ or $Y$, with probabilities given by the mixing proportions.
The distribution of the sum is 'more Normal' than either of the component distributions: adding two distributions tends to smooth out their special features.  The distribution of the mixture isn't; a mixture of two Normals is bimodal, because you get an observation from one or the other.
A: Adding another answer to the mix, in case it helps.
The random variable $Z = X + Y$ has a CDF of $P(Z\le z) = P(X+Y\le z)$.
On the other hand, if you have a mixture distribution $Z\sim 0.5X + 0.5Y$, then it
has PDF of $0.5f_X+0.5f_Y$, and CDF of $0.5F_X + 0.5F_Y$. That is, the first deals
with the actual event: we are asking, "what is the chance that the sum of a random
variable $X$ and $Y$ is a certain value?" The second has to do with likelihood.
We are asking, "what would happen if we added these likelihoods to construct
a new one?" As an example, if I had $X\sim N(0,1), Y\sim N(1000,1)$, then the
mixture distribution would look like two mountains. On the other hand, the
sum of the two would not have any peak at 0. Why? Because it is asking,
"What is the likelihood that $X+Y$ is near 0?" Well, since each time, $Y$ is drawn
from a distribution that is mean $1000$, the likelihood is actually quite low.
On the other hand, one way of thinking about the mixture distribution is that we are first picking
the distribution itself (either $X$ or $Y$), and then getting a value from it.
If that's the case, then the likelihood of getting something near $0$ is pretty high,
if you choose the variable $X$.
