Sorry if my question is trivial, but I am a bit confused about conditional probabilities when there are mixed variables, i.e., discrete and continuous. The problem is the following:
An experiment that is performed $n$ times can yield three possible outcomes each time it is done: left, middle, and right. For this, we use a trinomial distribution to calculate the probabilities associated with each outcome (say $p_1$, $p_2$, and $p_3$, respectively). If the outcome is ¨left¨, then the quantity $Y_1$ is won, where $Y_1$ is normally distributed with mean $\mu_1$ and standard deviation $\sigma_1$. If the outcome is ¨middle¨, then the quantity $Y_2$ is won, where $Y_2$ is normally distributed with mean $\mu_2$ and std $\sigma_2$. Finally, if the outcome is ¨right¨ then the quantity $Y_3$ is won, where $Y_3$ is normally distributed with mean $\mu_3$ and std $\sigma_3$.
What is the unconditional variance and expected value of Y (i.e., $E[Y]$,$\operatorname{Var}[Y]$)?
I understand that I would have to apply the law of iterated expectations and the law of total variance, but the formulation is unclear to me. Am I correct?
I would really appreciate any help that you can provide!
Best Answer
Yes.
Yes
Let $N\in\{1,2,3\}$ be the trinomial random variable .$N\sim\mathcal{Trinom}(p_1,p_2,p_3)$.
The Law of Iterated Expectation is:
$\qquad\begin{align}\mathsf E(Y) &= \mathsf E(\mathsf E(Y\mid N))\\&=p_1\,\mathsf E(Y_1)+p_2\,\mathsf E(Y_2)+p_3\,\mathsf E(Y_3)\\&=p_1\mu_1+p_2\mu_2+p_3\mu_3\end{align}$
Similarly, the Law of Iterated Variance is:
$\qquad\mathsf{Var}(Y) = \mathsf E(\mathsf {Var}(Y\mid N))+\mathsf{Var}(\mathsf E(Y\mid N))$
Now just apply the usual definitions for expectation and variance.