Let say, we have 2 IIDs $X, Y \sim N \left( 0, 1 \right) + \eta $
Now $\eta$ has a discrete distribution with values 0(-10) with probabilities 0.991 & 0.009 respectively.
Also assume that $Z = X+Y$.
I need to find the value of $z$ such that $P \left[ Z < z \right] = 1\%$
Is there any closed form solution available for the value of $z$?
Another way to express these quantities is to let $W$ be a standard Normal variable and $U$ be a Bernoulli$(0.009)$ variable. Both $X$ and $Y$ have the distribution of $W - 10U.$ Thus, $Z=X+Y$ has the distribution of (a) the sum of two iid standard Normal variables plus $-10$ times (b) the sum of two iid Bernoulli$(0.009)$ variables.
It is elementary that (a) has a Normal$(0,\sqrt{2})$ distribution and (b) has a Binomial$(0.009, 2)$ distribution. This latter takes on three values $0,1,2$ with chances $(1-p)^2, 2p(1-p),$ and $p^2,$ respectively (writing $p=0.009$). Subtracting $10$ times their value exhibits $Z$ as a mixture of three Normal variables with means $0$, $0-10(1)=-10,$ and $0-20(1)=-20.$ The mixture weights are $(1-p)^2, 2p(1-p),$ and $p^2,$ respectively.
Here is a plot of this mixture distribution (CDF) $F.$ I use a semi-log scale because there's a fairly large range of relevant probabilities:
The three Normal components centered at $0,-10,-20$ are apparent: these are the locations very close to the modes (where the slope of this plot is locally steepest). The red line shows the value $1\% = 0.01.$ The solution you seek is the value $z$ shown by the vertical gray line, situated where the red line intersects the graph.
Evidently, this solution is the zero of the function $z\to F(z)-1/100.$ Find it using any good univariate root finder. With double precision arithmetic you should obtain $z \approx -9.8006135477.$
This approach extends in an obvious manner to finding and working with distributions of the sums of any finite number of finite mixtures: it comes down to adding any pair of mixture components (such as $X$ and $Y$) and adding the discrete mixing variables (in this case, adding two iid copies of $U$).
