I saw this problem: you toss 10000 coins, 5000 silvers, 5000 golds. If you get head up on a gold coin. you get 4 dollars. If you get head up on a silver coin, you get 1 dollar. Given you get 14000 dollars, what's the expected number of gold coins with heads up?
The solution I saw says I just need to orthogonally project from $(5000, 5000)$ to $4x + y = 14000$. The interception is a close enough estimation.
Can someone explain why? It definitely makes sense but I can't rigorously prove it.
If we generalize the problem a bit, say we have a multivariate normal distribution $X\ \sim\ \mathcal{N}(\boldsymbol\mu, \boldsymbol\Sigma)$, and a set of linear constraints $Ax = b$, would a similar "trick" exist? What if the components are independent?
Firstly, mention some facts about conditional distribution of multivariate normal distribution.
Let $\boldsymbol{\xi}=(X_1,X_2,Z)^\top$ be multivariate normal distributed, i.e., \begin{gather*} \boldsymbol{\xi}\overset{d}{=} N(\boldsymbol{\mu},\boldsymbol{\Sigma} ),\\ \boldsymbol{\mu}=\begin{pmatrix} \mathsf{E}[X_1]\\ \mathsf{E}[X_2]\\\mathsf{E}[Z] \end{pmatrix} =\begin{pmatrix} \mu_1\\ \mu_2\\ \mu_Z \end{pmatrix},\quad \boldsymbol{\Sigma}=\begin{pmatrix} \sigma_{11}&\sigma_{12}&\sigma_{1Z}\\\sigma_{21}&\sigma_{22}&\sigma_{2Z}\\ \sigma_{Z1}&\sigma_{Z2}&\sigma_{ZZ} \end{pmatrix}. \end{gather*} Now the conditional distribution of $(X_1,X_2)^\top$ given $Z=z$ is normal distributed as following(please refer to T. W. Anderson, An Introduction to Multivariate Statistical Analysis, 3rd Ed. John Wiley & Sons, Inc. (2003). \$ 2.5.1. p.33.): \begin{gather*} \begin{pmatrix} X_{1|Z} \\ X_{2|Z} \end{pmatrix} \overset{d}{=} N(\overline{\boldsymbol{\mu}},\overline{\boldsymbol{\Sigma}} ),\tag{1}\\ \overline{\boldsymbol{\mu}}=\begin{pmatrix} \mu_{1|Z}\\ \mu_{2|Z} \end{pmatrix} =\begin{pmatrix} \mu_1+\dfrac{\sigma_{1Z}}{\sigma_{ZZ}}(z-\mu_Z) \\ \mu_2+\dfrac{\sigma_{2Z}}{\sigma_{ZZ}}(z-\mu_Z) \end{pmatrix},\tag{2}\\ \overline{\boldsymbol{\Sigma}} =\begin{pmatrix} \sigma_{11|Z}&\sigma_{12|Z}\\ \sigma_{21|Z}&\sigma_{22|Z} \end{pmatrix},\quad \sigma_{ij|Z}=\sigma_{ij}-\dfrac{\sigma_{iZ}\sigma_{jZ}}{\sigma_{ZZ}},\quad i,j=1,2. \tag{3} \end{gather*}
Now turn to your problem. Let $X_1(X_2 \;\text{resp.}) $ be the number of you get head up in toss 5000 gold(silver resp.) coin and $Z=4X_1+X_2$. Then $X,Y$ are mutual independent and both are binomial distributed, $X(Y)\overset{d}{=} b(5000, \frac12)$. Asymptotically, $X(Y)\overset{d}{=}N(2500,1250)$ and \begin{gather*} \begin{pmatrix} X_1\\ X_2 \\Z \end{pmatrix} \overset{d}{=} N(\boldsymbol{a},\boldsymbol{S}),\\ \boldsymbol{a}=\begin{pmatrix} 2500\\2500\\12500 \end{pmatrix},\qquad \boldsymbol{S}=\begin{pmatrix} 1250&0&5000\\0&1250&1250\\5000&1250&21250 \end{pmatrix}. \end{gather*} Hence, from (1)-(3), the conditional distribution of $(X_1,X_2)^\top$ given $Z=14000$ is $N(\boldsymbol{m},\boldsymbol{T})$, where \begin{equation*} \boldsymbol{m}=\begin{pmatrix} 2500+\dfrac{6000}{17}\\ 2500+\dfrac{1250}{17}\end{pmatrix},\qquad \boldsymbol{T}=\begin{pmatrix} \dfrac{20000}{17}&\dfrac{5000}{17}\\ -\dfrac{5000}{17}&-\dfrac{1250}{17} \end{pmatrix}. \end{equation*}