Return to Question

Let's say you have a set of order statistics $ X_{(1)}, \dots, X_{(N)} $ drawn from a discrete uniform distribution $ \text{unif}(1,S) $. If you choose $ X_{(n_1)}, \dots, X_{(n_k)} $ from this set, how would you find the distribution $ Y = \sum_{j=1}^k X_{(n_k)} $?

As an example to make it clearer, I've been trying to develop a tabletop RPG system and I'm thinking that the player could level up their stats in a few different ways. One way I'm imagining is that players can roll--for example--4 d6's and then must start off with $ X_{(1)} + X_{(2)} $, then being able to upgrade to something like $ X_{(2)} + X_{(3)} $. I'd like to work this out in general to see if it's balanced, but given that $ X_{(1)}, \dots, X_{(k)} $ are not i.i.d., this has been difficult.

[Edit]

I have still been considering this question, and I've been able to make some slight progress by considering a transformation of the sample space. In particular, let

$$ S = \{ X \in \textbf{N}^K : 1 \leq X_1 \leq \dots \leq X_K \leq T \} $$

If $ m_Y(r) $ counts the multiplicity of $ r $ in the vector $ Y $ (e.g., $ m_{(1,3,3)}(3) = 2 $), the p.m.f. of this distribution is,

$$ f(Y) = \frac{1}{T^K} \binom{K}{m_Y(1), \dots, m_Y(T)} $$

Questions about order statistics from the original distribution then become questions about the marginal probabilities of a given index. E.g., something like $ \Pr(X^{(1)} = 2) $ becomes $ \Pr(Y_2 = 2) $. Then the marginal distribution of $ Y_i $ is

$$ f_{Y_i}(z) &= \sum_{\substack{y \in S\\ y_i = z}} f(y) $$

This helps a bit, and it's more of an answer I had initially, but if there's a way of simplifying this sum I have been unable to find it.

Let's say you have a set of order statistics $ X_{(1)}, \dots, X_{(N)} $ drawn from a discrete uniform distribution $ \text{unif}(1,S) $. If you choose $ X_{(n_1)}, \dots, X_{(n_k)} $ from this set, how would you find the distribution $ Y = \sum_{j=1}^k X_{(n_k)} $?

As an example to make it clearer, I've been trying to develop a tabletop RPG system and I'm thinking that the player could level up their stats in a few different ways. One way I'm imagining is that players can roll--for example--4 d6's and then must start off with $ X_{(1)} + X_{(2)} $, then being able to upgrade to something like $ X_{(2)} + X_{(3)} $. I'd like to work this out in general to see if it's balanced, but given that $ X_{(1)}, \dots, X_{(k)} $ are not i.i.d., this has been difficult.

[Edit]

I have still been considering this question, and I've been able to make some slight progress by considering a transformation of the sample space. In particular, let

$$ S = \{ X \in \textbf{N}^K : 1 \leq X_1 \leq \dots \leq X_K \leq T \} $$

If $ m_Y(r) $ counts the multiplicity of $ r $ in the vector $ Y $ (e.g., $ m_{(1,3,3)}(3) = 2 $), the p.m.f. of this distribution is,

$$ f(Y) = \frac{1}{T^K} \binom{K}{m_Y(1), \dots, m_Y(T)} $$

Questions about order statistics from the original distribution then become questions about the marginal probabilities of a given index. E.g., something like $ \Pr(X^{(1)} = 2) $ becomes $ \Pr(Y_2 = 2) $. Then the marginal distribution of $ Y_i $ is

$$ f_{Y_i}(z) = \sum_{\substack{y \in S\\ y_i = z}} f(y) $$

This helps a bit, and it's more of an answer I had initially, but if there's a way of simplifying this sum I have been unable to find it.

Let's say you have a set of order statistics $ X_{(1)}, \dots, X_{(N)} $ drawn from a discrete uniform distribution $ \text{unif}(1,S) $. If you choose $ X_{(n_1)}, \dots, X_{(n_k)} $ from this set, how would you find the distribution $ Y = \sum_{j=1}^k X_{(n_k)} $?

As an example to make it clearer, I've been trying to develop a tabletop RPG system and I'm thinking that the player could level up their stats in a few different ways. One way I'm imagining is that players can roll--for example--4 d6's and then must start off with $ X_{(1)} + X_{(2)} $, then being able to upgrade to something like $ X_{(2)} + X_{(3)} $. I'd like to work this out in general to see if it's balanced, but given that $ X_{(1)}, \dots, X_{(k)} $ are not i.i.d., this has been difficult.

[Edit]

I have still been considering this question, and I've been able to make some slight progress by considering a transformation of the sample space. In particular, let

$$ S = \{ X \in \textbf{N}^K : 1 \leq X_1 \leq \dots \leq X_K \leq T \} $$

If $ m_Y(r) $ counts the multiplicity of $ r $ in the vector $ Y $ (e.g., $ m_{(1,3,3)}(3) = 2 $), the p.m.f. of this distribution is,

$$ f(Y) = \frac{1}{T^K} \binom{K}{m_Y(1), \dots, m_Y(T)} $$

Questions about order statistics from the original distribution then become questions about the marginal probabilities of a given index. E.g., something like $ \Pr(X^{(1)} = 2) $ becomes $ \Pr(Y_2 = 2) $. Then the marginal distribution of $ Y_i $ is

$$ \begin{align*} f_{Y_i}(z) &= \sum_{\substack{y \in S\\ y_i = z}} f(y) \\ &= \sum_{|a| = K} \binom{K}{a} \end{align*} $$

using multi-index notation with $ a = (a_1, \dots, a_T) $ such that $ a_i \geq 0 $ for $ i \neq z $ and $ a_z \geq 1 $.

This helps a bit, and it's more of an answer I had initially, but if there's a way of simplifying this sum I have been unable to find it.

Let's say you have a set of order statistics $ X_{(1)}, \dots, X_{(N)} $ drawn from a discrete uniform distribution $ \text{unif}(1,S) $. If you choose $ X_{(n_1)}, \dots, X_{(n_k)} $ from this set, how would you find the distribution $ Y = \sum_{j=1}^k X_{(n_k)} $?

As an example to make it clearer, I've been trying to develop a tabletop RPG system and I'm thinking that the player could level up their stats in a few different ways. One way I'm imagining is that players can roll--for example--4 d6's and then must start off with $ X_{(1)} + X_{(2)} $, then being able to upgrade to something like $ X_{(2)} + X_{(3)} $. I'd like to work this out in general to see if it's balanced, but given that $ X_{(1)}, \dots, X_{(k)} $ are not i.i.d., this has been difficult.

[Edit]

I have still been considering this question, and I've been able to make some slight progress by considering a transformation of the sample space. In particular, let

$$ S = \{ X \in \textbf{N}^K : 1 \leq X_1 \leq \dots \leq X_K \leq T \} $$

If $ m_Y(r) $ counts the multiplicity of $ r $ in the vector $ Y $ (e.g., $ m_{(1,3,3)}(3) = 2 $), the p.m.f. of this distribution is,

$$ f(Y) = \frac{1}{T^K} \binom{K}{m_Y(1), \dots, m_Y(T)} $$

Questions about order statistics from the original distribution then become questions about the marginal probabilities of a given index. E.g., something like $ \Pr(X^{(1)} = 2) $ becomes $ \Pr(Y_2 = 2) $. Then the marginal distribution of $ Y_i $ is

$$ f_{Y_i}(z) &= \sum_{\substack{y \in S\\ y_i = z}} f(y) $$

This helps a bit, and it's more of an answer I had initially, but if there's a way of simplifying this sum I have been unable to find it.

Let's say you have a set of order statistics $ X_{(1)}, \dots, X_{(N)} $ drawn from a discrete uniform distribution $ \text{unif}(1,S) $. If you choose $ X_{(n_1)}, \dots, X_{(n_k)} $ from this set, how would you find the distribution $ Y = \sum_{j=1}^k X_{(n_k)} $?

As an example to make it clearer, I've been trying to develop a tabletop RPG system and I'm thinking that the player could level up their stats in a few different ways. One way I'm imagining is that players can roll--for example--4 d6's and then must start off with $ X_{(1)} + X_{(2)} $, then being able to upgrade to something like $ X_{(2)} + X_{(3)} $. I'd like to work this out in general to see if it's balanced, but given that $ X_{(1)}, \dots, X_{(k)} $ are not i.i.d., this has been difficult.

[Edit]

I have still been considering this question, and I've been able to make some slight progress by considering a transformation of the sample space. In particular, let

$$ S = \{ X \in \textbf{N}^K : 1 \leq X_1 \leq \dots \leq X_K \leq T \} $$

that the original question about $ X^{(i)} = x $ from $ X^1, \dots, X^K $ drawn from $ \text{unif}(1, T) $ (where the superscript is an index) instead becomes a question about $ Y_i = x $ for a random vector $ Y $ from $ S $. Then, if $ m_Y(r) $ counts the multiplicity of $ r $ in the vector $ Y $ (e.g., $ m_{(1,3,3)}(3) = 2 $), the p.m.f. of this distribution is,

$$ f(Y) = \frac{1}{T^K} \binom{K}{m_Y(1), \dots, m_Y(T)} $$

Questions about order statistics then become questions about the marginal probabilities of a given index. E.g., something like $ \Pr(X^{(1)} = 2) $ becomes $ \Pr(Y_2 = 2) $. Then the marginal distribution of $ Y_i $ is

$$ \begin{align*} f_{Y_i}(z) &= \sum_{\substack{y \in S\\ y_i = z}} f(y) \\ &= \sum_{|a| = K} \binom{K}{a} \end{align*} $$

using multi-index notation with $ a = (a_1, \dots, a_T) $ such that $ a_i \geq 0 $ for $ i \neq j $ and $ a_y \geq 1 $.

This helps a bit, and it's more of an answer I had initially, but if there's a way of simplifying this sum I have been unable to find it.

Let's say you have a set of order statistics $ X_{(1)}, \dots, X_{(N)} $ drawn from a discrete uniform distribution $ \text{unif}(1,S) $. If you choose $ X_{(n_1)}, \dots, X_{(n_k)} $ from this set, how would you find the distribution $ Y = \sum_{j=1}^k X_{(n_k)} $?

As an example to make it clearer, I've been trying to develop a tabletop RPG system and I'm thinking that the player could level up their stats in a few different ways. One way I'm imagining is that players can roll--for example--4 d6's and then must start off with $ X_{(1)} + X_{(2)} $, then being able to upgrade to something like $ X_{(2)} + X_{(3)} $. I'd like to work this out in general to see if it's balanced, but given that $ X_{(1)}, \dots, X_{(k)} $ are not i.i.d., this has been difficult.

[Edit]

I have still been considering this question, and I've been able to make some slight progress by considering a transformation of the sample space. In particular, let

$$ S = \{ X \in \textbf{N}^K : 1 \leq X_1 \leq \dots \leq X_K \leq T \} $$

If $ m_Y(r) $ counts the multiplicity of $ r $ in the vector $ Y $ (e.g., $ m_{(1,3,3)}(3) = 2 $), the p.m.f. of this distribution is,

$$ f(Y) = \frac{1}{T^K} \binom{K}{m_Y(1), \dots, m_Y(T)} $$

Questions about order statistics from the original distribution then become questions about the marginal probabilities of a given index. E.g., something like $ \Pr(X^{(1)} = 2) $ becomes $ \Pr(Y_2 = 2) $. Then the marginal distribution of $ Y_i $ is

$$ \begin{align*} f_{Y_i}(z) &= \sum_{\substack{y \in S\\ y_i = z}} f(y) \\ &= \sum_{|a| = K} \binom{K}{a} \end{align*} $$

using multi-index notation with $ a = (a_1, \dots, a_T) $ such that $ a_i \geq 0 $ for $ i \neq z $ and $ a_z \geq 1 $.

This helps a bit, and it's more of an answer I had initially, but if there's a way of simplifying this sum I have been unable to find it.