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

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 $X$.
