Suppose $n = 18, (n_1, n_2, n_3)$ follow multinomial distribution with $p_i = (0.6, 0.3, 0.1)$, let $v = n_1 + 3 n_2 + 10 n_3$
Then what is the variance of $v$ given that $n_3 = 2$? i.e. Var$(v \mid n_3 = 2)$
Moreover, if $n \sim$ Pois$(18)$, what is the variance of $v$ given that $n_3 = 2$?
I calculated the varinace of $v$ (given $n = 18$): \begin{align*} \text{Var} (v) &= \text{Var} (n_1 + 3 n_2 + 10 n_3) \\ &= \text{Var} (n_1) + 9 \text{Var} (n_2) + 100 \text{Var} (n_3) - 2[3 \text{Cov} (n_1, n_2) + 3 \cdot 10 \text{Cov} (n_2, n_3) + 10 \text{Cov} (n_1, n_3)] \\ &= n (0.6 \cdot 0.4 + 9 \cdot 0.3 \cdot 0.7 + 100 \cdot 0.1 \cdot 0.9) - 2n [3 \cdot 0.6 \cdot 0.3 + 30 \cdot 0.3 \cdot 0.1 + 10 \cdot 0.6 \cdot 0.1] \\ &= 126.9 \end{align*}
How can I find the conditional varianace given that $n_3 = 2$?
Let's think about the distribution of $(n_1, n_2)$ given $n_3 = 2$ and $n = 18$. This means $n_1 + n_2 = 16$, so $n_1$ must be binomial, and our intuition suggests that since $p_1 = 0.6 = 2p_2$, $$\Pr[n_1 = x \mid n_3 = 2] = \binom{16}{x} (2/3)^x (1/3)^{16-x}.$$ But how do we show this? We note by the definition of conditional probability, $$\begin{align} \Pr[(n_1, n_2) = (x, 16-x) \mid n_3 = 2] &= \frac{\Pr[(n_1, n_2, n_3) = (x, 16-x, 2)]}{\Pr[n_3 = 2]} \\ &= \frac{\frac{18!}{x! (16-x)! 2!} (0.6)^x (0.3)^{16-x} (0.1)^2}{\binom{18}{2} (0.1)^2 (0.9)^{18-2}} \\ &= \frac{16!}{x! (16-x)!} \left(\frac{0.6}{0.9}\right)^x \left(\frac{0.3}{0.9}\right)^{16-x} \\ &= \binom{16}{x} (2/3)^x (1/3)^{16-x}, \end{align}$$ as claimed.
Therefore, the conditional variance of $v = n_1 + 3n_2 + 10n_3$ given $n_3 = 2$ is simply
$$\operatorname{Var}[n_1 + 3(16-n_1) + 10n_3 \mid n_3 = 2] = 4 \operatorname{Var}[n_1 \mid n_3 = 2] = 4 (16)(2/3)(1/3) = \frac{128}{9}.$$