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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!

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  • $\begingroup$ Try and write down the formulas for the mean and variance of $Y$. $\endgroup$
    – Yuval Peres
    Jul 13, 2022 at 15:44
  • $\begingroup$ @YuvalPeres Is it something like this? E[Y]=E[Y|X="left"]Pr(X="left")+E[Y|X="middle"]Pr(X="middle")+E[Y|X="right"]Pr(X="right")=𝜇1𝑝1+𝜇2𝑝2+𝜇3𝑝3 $\endgroup$
    – angelavtc
    Jul 13, 2022 at 16:14

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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?

Yes.

Is it something like this? E[Y]=E[Y|X="left"]Pr(X="left")+E[Y|X="middle"]Pr(X="middle")+E[Y|X="right"]Pr(X="right")=𝜇1𝑝1+𝜇2𝑝2+𝜇3𝑝

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.

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  • $\begingroup$ Thank you very much, I have tried to do the variance, could you please tell me if my computation is correct? $\operatorname{Var}[Y]=E[\operatorname{Var}[Y|N])]+\operatorname{Var}[E[Y|N]]$ where $\operatorname{Var}[Y|N]=\operatorname{Var}[\sum\limits_{i=1}^{3}y_{i}|N]=\sum\limits_{n=1}^{3}\sigma_{i}^{2} $ then $\operatorname{Var}[Y]=E[\sum\limits_{n=1}^{3}\sigma_{i}^{2}]+\operatorname{Var}[\sum\limits_{n=1}^{3}p_{i}\mu_i]=\sum\limits_{n=1}^{3}\sigma_{i}^{2}+\sum\limits_{n=1}^{3}p_{i}^{2}\mu_{i}^{2}$ $\endgroup$
    – angelavtc
    Jul 14, 2022 at 10:08
  • $\begingroup$ Am I right with $\operatorname{Var}[Y|N]=\operatorname{Var}[\sum\limits_{i=1}^{3}y_{i}|N]=\sum\limits_{n=1}^{3}\sigma_{i}^{2} $? or is it $\operatorname{Var}[Y|N]=\operatorname{Var}[\sum\limits_{i=1}^{3}y_{i}|N]=\sum\limits_{n=1}^{3}p_i^{2}\sigma_{i}^{2} $ I am kind of confused :/ Thank you! $\endgroup$
    – angelavtc
    Jul 14, 2022 at 11:47
  • $\begingroup$ It is that $\mathsf{Var}(Y\mid N)=\sigma^2_N$, so therefore $\mathsf E(\mathsf{Var}(Y\mid N))=\sum_{n=1}^3p_n\sigma^2_n$. Likewise, $\mathsf E(Y\mid N)=\mu_N$, so $\mathsf{Var}(\mathsf E(Y\mid N))=\ldots$ $\endgroup$
    – Graham Kemp
    Jul 14, 2022 at 12:58
  • $\begingroup$ I see, then we start with:\\ $\operatorname{Var}[Y|N]=\sigma_{N}^{2}$ and $E[Y|N]=\mu_N$\\ This means:\\ $E[\operatorname{Var}[Y|N]]=\sum\limits_{i=1}^{3}p_{i}\sigma_i^{2}$\\ Moreover: $Var[E[Y|N]]=\sum\limits_{i=1}^{3}p_{i}^{2}\mu_{i}^{2}$\\ Therefore, $Var[Y]=\sum\limits_{n=1}^{3}p_{i}\sigma_i^{2}+\sum\limits_{n=1}^{3}p_{i}^{2}\mu_{i}^{2}$ Is this ok? $\endgroup$
    – angelavtc
    Jul 14, 2022 at 15:55
  • $\begingroup$ No. @angelavtc : $\mathsf {Var}(\mu_N) ~{=\mathsf E(\mu_N^2)-\mathsf E(\mu_N)^2\\ =\sum_{n=1}^3 p_n\mu_n^2 -\left(\sum_{n=1}^3 p_n\mu_n\right)^2}$ $\endgroup$
    – Graham Kemp
    Jul 14, 2022 at 22:25

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