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Let $X,Y,Z$ be independent and identically distributed Bernoulli Random Variables with parameter $p$. I am looking for a way to compute the Comulative Distribution Funktion of $U=(1−X)Y+ XZ$

I have only come across easier examples where 2 simple case differentiations were enough but I can't see a simple solution for this one. Am I missing something?

Bonus points for $\mathbb{E}[U]$ and $Var[U]$

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    Bonus points for showing a little effort. – 2017-01-02

2 Answers 2

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So $U=Y$ if $X=0$ and $U=Z$ if $X=1$. So $U\in\{0,1\}$.

Thus $P(U=0)=P(X=0,Y=0)+P(X=1,Z=0)=(1-p)^2+p(1-p)=1-p$ and therefore $P(U=1)=p$.

So $E(U)=p$ and $Var(U)=E(U^2)-(E(U))^2=p-p^2=p(1-p)$.

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if we condition on $X$ we can use the independence to get the following for $u \in \mathbb R$: \begin{align} \mathbb P (U \leq u) &= \mathbb P ( U \leq u \mid X=1) \cdot \mathbb P (X=1)+\mathbb P ( U \leq u \mid X=0) \cdot \mathbb P (X=0) \\ &= \mathbb P (Z \leq u \mid X=1) \cdot p+\mathbb P (Y \leq u \mid X=0) \cdot (1-p)\\ &=\mathbb P (Z \leq u) \cdot p+\mathbb P (Y \leq u) \cdot (1-p)\\ &=\begin{cases} 0\cdot p + 0 \cdot (1-p) = 0, & \text{if} \ u<0, \\ p\cdot p + p\cdot (1-p) = p, & \text{if} \ u \in [0,1), \\ 1\cdot p + 1 \cdot (1-p) = 1, &\text{if} \ u\geq 1. \end{cases} \end{align} In particular $U$ has the same distribution as $X,Y$ and $Z$. Hence $$ \mathbb{E}[U]=\mathbb{E}[X]=p, \quad Var(U)=Var(X)=p\cdot(1-p). $$