Notation
- # categories $= c$.
- # trials $= t$.
- Side $i$ = $si$.
- Random vector $= S = \left[S_1\;S_2\;\ldots\;S_c\right]^T$.
- # occurrences of $si$ vector $= s = \left[s1\;s2\;\ldots\;sc\right]^T$.
- Probability of occurrence of $si$ vector $= p = \left[p_{s1}\;p_{s2}\;\ldots\;p_{sc}\right]^T$.
- $p_{s1} + p_{s2} + \cdots + p_{sc} = 1$.
- $s1 + s2 + \cdots + sc = t$.
- Pmf of $S = P\left[S = s\right]$.
Games
- g1: An ideal 2 sided die is simulated using a standard die by assigning faces w/ dots 1 through 3 & 4 through 6 to $s1$ & $s2$, respectively. The 2 sided die is tossed 7 times & the # times $s1$ & $s2$ occur are recorded.
- g2: Same concept as in g1, accept faces w/ dots 1 through 3 have micro holes etched into them s.t. $p_{s2} = 18/30$.
- g3: Same concept as in g1, accept w/ the standard die, i.e., $p_{s1} = p_{s2} = \cdots = p_{s6} = 5/30$.
- g4: Same concept as in g2, accept w/ the standard die, i.e., $p_{s4} = p_{s5} = p_{s6} = 6/30$.
- g5: Same concept as in g2, accept the micro holes are filled w/ $0.07$ kg of a material, which evaporates @ $0.01$ kg/s upon being sprayed w/ an activator, s.t. $p_{s2} = 15/30$ for the 1st toss. Immediately after being sprayed, the outcomes of the tosses are measured every second.
- g6: Same concept as in g5, accept w/ the standard die, i.e., $p_{s4} = p_{s5} = p_{s6} = 5/30$ for the 1st toss.
Questions
- q1: Find pmf & evaluate when $s = \left[2\;5\right]^T$.
- q2: q1.
- q3: Find pmf & evaluate when $s = \left[0\;2\;1\;1\;0\;3\right]^T$.
- q4: q3.
- q5: q1.
- q6: q3.
Answers
- a1: $S$ ~ Binomial distribution.
- $P\left[S = s\right] = t!\prod_{k = 1}^c \frac{p_{sk}^{sk}}{sk!} = 7!\prod_{k = 1}^2 \frac{(15/30)^{sk}}{sk!} = \frac{7!(15/30)^2(15/30)^5}{2!5!}$
$\Longrightarrow P\left[S = s\right] = 21/128$.
- $P\left[S = s\right] = t!\prod_{k = 1}^c \frac{p_{sk}^{sk}}{sk!} = 7!\prod_{k = 1}^2 \frac{(15/30)^{sk}}{sk!} = \frac{7!(15/30)^2(15/30)^5}{2!5!}$
- a2: $S$ ~ Binomial distribution.
- $P\left[S = s\right] = t!\prod_{k = 1}^c \frac{p_{sk}^{sk}}{sk!} = 7!\prod_{k = 1}^2 \frac{p_{sk}^{sk}}{sk!} = \frac{7!(12/30)^2(18/30)^5}{2!5!}$
$\Longrightarrow P\left[S = s\right] = 20412/78125$.
- $P\left[S = s\right] = t!\prod_{k = 1}^c \frac{p_{sk}^{sk}}{sk!} = 7!\prod_{k = 1}^2 \frac{p_{sk}^{sk}}{sk!} = \frac{7!(12/30)^2(18/30)^5}{2!5!}$
- a3: $S$ ~ Multinomial distribution.
- $P\left[S = s\right] = t!\prod_{k = 1}^c \frac{p_{sk}^{sk}}{sk!} = 7!\prod_{k = 1}^6 \frac{(5/30)^{sk}}{sk!} = \frac{7!(5/30)^2(5/30)(5/30)(5/30)^3}{0!2!1!1!0!3!}$
$\Longrightarrow P\left[S = s\right] = 35/23328$.
- $P\left[S = s\right] = t!\prod_{k = 1}^c \frac{p_{sk}^{sk}}{sk!} = 7!\prod_{k = 1}^6 \frac{(5/30)^{sk}}{sk!} = \frac{7!(5/30)^2(5/30)(5/30)(5/30)^3}{0!2!1!1!0!3!}$
- a4: $S$ ~ Multinomial distribution.
- $P\left[S = s\right] = t!\prod_{k = 1}^c \frac{p_{sk}^{sk}}{sk!} = 7!\prod_{k = 1}^6 \frac{p_{sk}^{sk}}{sk!} =
\frac{7!(4/30)^2(4/30)(6/30)(6/30)^3}{0!2!1!1!0!3!}$
$\Longrightarrow P\left[S = s\right] = 224/140625$.
- $P\left[S = s\right] = t!\prod_{k = 1}^c \frac{p_{sk}^{sk}}{sk!} = 7!\prod_{k = 1}^6 \frac{p_{sk}^{sk}}{sk!} =
\frac{7!(4/30)^2(4/30)(6/30)(6/30)^3}{0!2!1!1!0!3!}$
- a5: $S$ ~ Poisson's binomial distribution.
- $P\left[\left[S_1\;S_2\right]^T = \left[s1\;s2\right]^T\right] = P\left[S_1 = s1, S_2 = s2\right] = P\left[S_1 = s1\right] = P\left[S_2 = s2\right]$.
- $p_{s1}$ & $p_{s2}$ are vectors now: $p_{s1} = \left[p_{s1_1}\;p_{s1_2}\;\ldots\;p_{s1_t}\right]^T, p_{s2} = \left[p_{s2_1}\;p_{s2_2}\;\ldots\;p_{s2_t}\right]^T$.
- $P\left[S = s\right] = \frac{1}{t + 1}\sum_{i = 0}^t \left\{\exp\left(\frac{-j2\pi i s}{t + 1}\right) \prod_{k = 1}^t \left\{p_{s2_k}\left(\exp\left(\frac{j2\pi i}{t + 1}\right) - 1\right) + 1\right\}\right\}$
$= \frac{1}{8}\sum_{i = 0}^7 \left\{\exp\left(\frac{-j5\pi i}{4}\right) \prod_{k = 1}^7 \left\{\left(\frac{0.5k + 14.5}{30}\right)\left(\exp\left(\frac{j\pi i}{4}\right) - 1\right) + 1\right\}\right\}$
$\Longrightarrow P\left[S_2 = 5\right] = 53529/250000$.
- a6: $S$ ~ Poisson's multinomial distribution.
- I understand this enough to write a q about it, but not enough to solve it.
- This q was the only 1 on SE that was relatable but the answer was incorrect.
- PLEASE help me find the exact solution; not an approximation.