0

My answer still unaddressed for some unknown reasons. Given the constants $\{a,b,c,d,e,f$}, I want to compute the conditional mean $\text{E}[Z|S_1,S_2]$ and the conditional variance $\text{Var}[Z|S_1,S_2]$, with:

$Z=a+bX_1+cX_2+dY_1+eY_2+fY_3$

Is the following true?

$\text{E}[Z|S_1,S_2]=a+b\text{E}[X_1|S_1,S_2]+c\text{E}[X_2|S_1,S_2]$

and

$\text{Var}[Z|S_1,S_2]=b^2\text{Var}[X_1|S_1,S_2]+c^2\text{Var}[X_2|S_1,S_2]+d^2\sigma_{Y_1}^2+e^2\sigma_{Y_2}^2+f^2\sigma_{Y_3}^2+bc\text{Cov}[X_1,X_2|S_1,S_2]+2de\text{Cov}[Y_1,Y_2]+2df\text{Cov}[Y_1,Y_3]+2ef\text{Cov}[Y_2,Y_3]$

where $\text{Cov}[X_1,X_2|S_1,S_2]=\text{E}[X_1X_2|S_1,S_2]-\text{E}[X_1|S_1,S_2]\text{E}[X_2|S_1,S_2]$

Assume $S_1=X_1+\epsilon_{X_1}, S_2=X_2+\epsilon_{X_2}$ (where $\epsilon_{X_1}\sim \mathcal N(0,\sigma_{\epsilon_{X_1}}^2)$ and $\epsilon_{X_2}\sim \mathcal N(0,\sigma_{\epsilon_{X_2}}^2)$) and the following joint distributions:

$\begin{pmatrix} X_1 \\ X_2 \end{pmatrix}$ $\sim \mathcal N$ $\bigg(\begin{pmatrix} \mu_{X_1} \\ \mu_{X_2} \end{pmatrix}, \begin{pmatrix} \sigma_{X_1}^2 & \rho_{X_1X_2}\sigma_{X_1}\sigma_{X_2}\\ * & \sigma_{X_2}^2 \end{pmatrix}\bigg)$

$\begin{pmatrix} Y_1 \\ Y_2 \\ Y_3 \end{pmatrix}$ $\sim \mathcal N$ $\Bigg(\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} \sigma_{Y_1}^2 & \rho_{Y_1Y_2}\sigma_{Y_1}\sigma_{Y_2} & \rho_{Y_1Y_3}\sigma_{Y_1}\sigma_{Y_3}\\ * & \sigma_{Y_2}^2 & \rho_{Y_2Y_3}\sigma_{Y_2}\sigma_{Y_3}\\ * & * & \sigma_{Y_3}^2 \end{pmatrix}\Bigg)$

Assume also that $(X_1,X_2)$ and $(S_1,S_2)$ are independent from $(Y_1,Y_2,Y_3)$.

  • 3
    If you carefully studied the answer in your previous question, at least for the conditional variance part, you should be able to derive the answer by yourself. The idea does not change at all between these two questions, even after you added more variates. – Zhanxiong Apr 06 '23 at 14:33
  • 7
    Hi anonymous. I know it can be frustrating when your queries get closed. But ranting in all caps is not an answer. Have patience. Take things slowly. Explain clearly why your query is not the same as the cited one. Everyone is here to help each other. No need to heat up. – User1865345 Apr 06 '23 at 16:26
  • How does this differ from your other question? If you think the current form is clearer, perhaps it would be easier to have it in the original and argue for it's reopening – Firebug Apr 12 '23 at 11:04
  • 1
    I deleted my previous question. In this current question I propose an answer myself but want to verify whether my understanding is correct. – anonymous Apr 12 '23 at 11:35
  • Is the following true?

    $\text{E}[Z|S_1,S_2]=a+b\text{E}[X_1|S_1,S_2]+c\text{E}[X_2|S_1,S_2]$

    and

    $\text{Var}[Z|S_1,S_2]=b^2\text{Var}[X_1|S_1,S_2]+c^2\text{Var}[X_2|S_1,S_2]+d^2\sigma_{Y_1}^2+e^2\sigma_{Y_2}^2+f^2\sigma_{Y_3}^2+bc\text{Cov}[X_1,X_2|S_1,S_2]+2de\text{Cov}[Y_1,Y_2]+2df\text{Cov}[Y_1,Y_3]+2ef\text{Cov}[Y_2,Y_3]$

    – anonymous Apr 12 '23 at 15:00
  • where $\text{Cov}[X_1,X_2|S_1,S_2]=\text{E}[X_1X_2|S_1,S_2]-\text{E}[X_1|S_1,S_2]\text{E}[X_2|S_1,S_2]$ – anonymous Apr 12 '23 at 15:02
  • Is the following true?

    $\text{E}[Z|S_1,S_2]=a+b\text{E}[X_1|S_1,S_2]+c\text{E}[X_2|S_1,S_2]$

    and

    $\text{Var}[Z|S_1,S_2]=b^2\text{Var}[X_1|S_1,S_2]+c^2\text{Var}[X_2|S_1,S_2]+d^2\sigma_{Y_1}^2+e^2\sigma_{Y_2}^2+f^2\sigma_{Y_3}^2+bc\text{Cov}[X_1,X_2|S_1,S_2]+2de\text{Cov}[Y_1,Y_2]+2df\text{Cov}[Y_1,Y_3]+2ef\text{Cov}[Y_2,Y_3]$

    – anonymous Apr 12 '23 at 15:12
  • where $\text{Cov}[X_1,X_2|S_1,S_2]=\text{E}[X_1X_2|S_1,S_2]-\text{E}[X_1|S_1,S_2]\text{E}[X_2|S_1,S_2]$ – anonymous Apr 12 '23 at 15:12

0 Answers0