Random Variable Transformation normal/binomial

Question

I have the following problem I can not solve:

We have two indipendent random Variables given by: $$ X \sim N_{(0,1)} $$ and $$ Y_p \sim B_{(1,p)} $$ Now I want to show, that $Z_p \sim N_{(0,1)}$ $\forall p \in (0,1)$ , with $$ Z_p = (-1)^{Y_p}\cdot X $$ Additionally there is a Hint whoch says, that: $$ \mathbb{P}(A) = \mathbb{P}(A \cap \{ Y_p = 1\}) + \mathbb{P}(A \cap \{ Y_p = 0\}) $$

I know that a transformation is given by: $$ F_Z(z) = \mathbb{P}(Z_p < z) = \mathbb{P}((-1)^{Y_p}\cdot X < z) $$ And this I somehow get my boundarys for the integration. But I cannot figure out how this is done.

Answer 1

I preassume that $X$ and $Y_p$ are independent.

For every Borel measurable $A$: $$\begin{aligned}P\left(Z_{p}\in A\right) & =P\left(Z_{p}\in A\mid Y_{p}=0\right)P\left(Y_{p}=0\right)+P\left(Z_{p}\in A\mid Y_{p}=1\right)P\left(Y_{p}=1\right)\\ & =P\left(X\in A\mid Y_{p}=0\right)\left(1-p\right)+P\left(-X\in A\mid Y_{p}=1\right)p\\ & =P\left(X\in A\right)\left(1-p\right)+P\left(-X\in A\right)p\\ & =P\left(X\in A\right)\left(1-p\right)+P\left(X\in A\right)p\\ & =P\left(X\in A\right) \end{aligned} $$

The third equality rests on independence and the fourth equality rests on the fact that $X$ and $-X$ have the same distribution.

Answer 2

Welcome to MSE! :-)

First note, that if $X \sim \mathrm{N}(0,1)$, that also $-X \sim \mathrm{N}(0,1)$, you can see this when noting, that the probability density function of $\mathrm{N}(0,1)$ is symmetric around $0$. Which to me would be an intuitive explanation of why this statement is true. However, this explanation works only, if $X$ and $Y_p$ are independent, hence I will assume that in the following.

Anyway, let's dig more into detail. Let $A \subseteq \mathbb{R}$ (measurable) and \begin{align*}\mathbb{P}(Z_p \in A) &= \mathbb{P}(\{Z_p \in A\} \cap \{Y_p = 1\}) + \mathbb{P}(\{Z_p \in A\} \cap \{Y_p = 0\}) \\ &= \mathbb{P}(\{-X \in A\} \cap \{Y_p = 1\}) + \mathbb{P}(\{X \in A\} \cap \{Y_p = 0\}).\end{align*} The last statement is true, since we know whether $Z_p = -X$ or $Z_p = X$, if we fix $Y_p$. Since $X$ and $Y_p$ are independent, we obtain \begin{align*}\mathbb{P}(\{-X \in A\} \cap \{Y_p = 1\}) &= p\mathbb{P}(\{-X \in A\}, \\ \mathbb{P}(\{X \in A\} \cap \{Y_p = 0\}) &= (1-p)\mathbb{P}(\{X \in A\}.\end{align*} The symmetry of $\mathrm{N}(0,1)$ implies that $\mathbb{P}(X \in A) = \mathbb{P}(-X \in A)$. Applying that first displayed formula, we obtain $$\mathbb{P}(Z_p \in A) = (p + 1 -p)\mathbb{P}(X \in A) = \mathbb{P}(X \in A) = \mathrm{N}(0,1)(A).$$ Hence, $Z_p \sim \mathrm{N}(0,1)$.