Let $X$ be a random variable that assumes the value 1 with probability $p$ and $-N$ with probability $1-p$.
$N$ is a Poisson random variable with $\lambda$ as parameter.
Solve:
a)the $\lambda$ value $/\mathbb{E}(X)=0$;
b)$Var(X)$;
c)Given the succession $\mathop{\{}X_i\mathop{\}}\limits_{i=1}^{n}$of random variables distributed as $X$;
Given $Y$ $= \sum_{i=1}^{M}$ with M Poisson's random variables of parameter $\beta$, independents from $X_i$;
Determine $\mathbb{E}(Y)$.
Here is my attempt:
a)${E}(X)=0$ $\implies$ $p-\lambda(1-p)=0\implies\lambda=p/(1-p)$
b)$Var(X)=\mathbb{E}(X^2)-E(X)^2=[p+\lambda^2(1-p)]-[p-\lambda(1-p)]^2$
c)I honestly don't know where to start ... any hints?
a) Looks good.
b) I think you've made a mistake; you've replaced $N$ with its expectation ($\lambda$), which doesn't work in the variance formula since $\mathbb E[N^2] \neq \mathbb E[N]^2$. Instead, consider the law of total variance: $$\operatorname{Var}(X) = \mathbb E \left[ \operatorname{Var} \left(X \mid N \right) \right] + \operatorname{Var} \left( \mathbb E[ X \mid N ] \right)$$ Inside the brackets and parentheses on the right side, you can compute epxressions involving $X$ by treating $N$ as a fixed number (and not a random variable). Then, when you proceed to the outer operators, you can account for the variance and expected value of $N$.
c) Consider Wald's identity: if $\{X_i\}$ are i.i.d., $K$ is a variable independent of $\{X_i\}$, and $X_i, K$ have finite expectation, then $$\mathbb E \left[ \sum_{i=1}^{K} X_i\right] = \mathbb E[K] \cdot \mathbb E[X_1].$$ The intuition of this formula is: suppose you roll a six-sided die and call its value $K$, then flip $K$ coins and count the number of heads. On average, you will flip $3.5$ coins, of which $0.5$ will be heads, leading to an expected value of $3.5 \cdot 0.5$ total heads.