We can predictively manipulate our inference regardless of what data we will get. If I decide to bug the person in the room one more time, I know beforehand that my best estimate would be N=k+1 . So I don't even need to ask the person anymore. I can already update my belief!

As Cliff noted, this isn't necessarily true. Assuming $k<N$ samples with $k$ different results $d_1,...,d_k$, you're always equally likely to get $d_{k+1}$ very close to one of the previous $k$ samples. This is true even for a certain range of $k>N$ values, as there's no guarantee that upon hitting $k=N$ requests you have been "exposed" to all $N$ options. The only thing that is guaranteed here is that for $K\ge N+1$ you have at least one cluster with two or more samples, due to the pigeonhole principle.

Taking no assumptions other than $N\le100$, you can solve this with Bayesian GMM and the Gibbs sampling algorithm. As you have known variance for the Gaussian models themselves and a strong assumption regarding the mixture probabilities. However we still take the probabilities vector $\pi$ as unknown so we don't limit ourselves to a fixed number of clusters (you'll see later why this works).

We denote $\theta=\{\pi,\{\mu_i\}_{i=1}^100\}$ our vector of parameters. An initial prior for $\pi$ would be $\pi\sim Dirichlet(\alpha_1,...,\alpha_{100})$. Of course $\forall i:\alpha_i>0, \pi_i\ge0, \sum_i\pi_i=1$. As you have a uniform assumption regarding the identities in the keystroke, you might want to take here $\alpha_1=...=\alpha_{100}$ and give it a high value (which reflects your high certainty in the Uniform manner).

Next the expectation vector with prior $\mu_i\sim U[1,10^6]$. Using the Bayes theorem we get $P(\mu|D)\propto P(D|\mu)P(\mu)$ (the denominator is a normalization constant which we can overlook for the moment). As the likelihood $P(D|\mu)$ is Gaussian and the prior $P(\mu)$ is uniform, the posterior $P(\mu|D)$ is a truncated Gaussian.

We start by choosing initial values for $\pi_i,\mu_i$ (for all $i$ values) and denote them $\pi_i^{(0)},\mu_i^{(0)}$. We then work iteratively, at each iteration $t+1$ we calculate the following steps, each step is calculated for all $k$ samples:

1. Responsibility $q_{ji}^{(t+1)}$ is the predicted probability of the sample $D_j$ belonging to the $i^{th}$ cluster after step $t$: $$q_{ji}^{(t+1)}=\frac{\pi_i^{(t)}\mathcal{N}(D_j;\mu_i^{(t)},1)}{\sum_{i=1}^N \pi_i^{(t)}\mathcal{N}(D_j;\mu_i^{(t)},1)}$$

2. Affiliation $$k_j^{(t+1)}\sim Multinomial \left(q_{j1}^{(t+1)},q_{j2}^{(t+1)},...,q_{jN}^{(t+1)}\right)$$

3. Cluster sizes: $$N_i^{(t+1)}=\sum_{j=1}^k\mathbb{I}\{k_j^{(t+1)}=j\}$$

4. Sampling the probabilities vector: $$(\pi^{(t+1)}|k^{(t+1)},D)\sim Dirichlet\left(N_1^{(t+1)}+\alpha_1,...,N_N^{(t+1)}+\alpha_N\right)$$

5. Calculating the expectations vector: Let $D^{i,(t+1)}$ be the subset of samples belonging to the $i^{th}$ cluster at this step: $$(\mu_i^{(t+1)}|k^{(t+1)},D^{i,(t+1)})=\frac{1}{N_i^{(t+1)}}\sum D^{i,(t+1)}$$

What happens eventually is that if the actual number of clusters is smaller that what we've initialized to, we'll get those clusters with 0 members and close to 0 mixture probability. You can also verify the number of clusters using the evidence function. Upon receiving a new sample $k+1$ you can re-run the whole process or (with some risks) initialize to the previous values.

Note that for the whole process, we have sticked with the basic assumptions and took nothing more.

We can predictively manipulate our inference regardless of what data we will get. If I decide to bug the person in the room one more time, I know beforehand that my best estimate would be N=k+1 . So I don't even need to ask the person anymore. I can already update my belief!

As Cliff noted, this isn't necessarily true. Assuming $k<N$ samples with $k$ different results $d_1,...,d_k$, you're always equally likely to get $d_{k+1}$ very close to one of the previous $k$ samples. This is true even for a certain range of $k>N$ values, as there's no guarantee that upon hitting $k=N$ requests you have been "exposed" to all $N$ options. The only thing that is guaranteed here is that for $K\ge N+1$ you have at least one cluster with two or more samples, due to the pigeonhole principle.

Taking no assumptions other than $N\le100$, you can solve this with Bayesian GMM and the Gibbs sampling algorithm. As you have known variance for the Gaussian models themselves and a strong assumption regarding the mixture probabilities. However we still take the probabilities vector $\pi$ as unknown so we don't limit ourselves to a fixed number of clusters (you'll see later why this works).

We denote $\theta=\{\pi,\{\mu_i\}_{i=1}^{100}\}$ our vector of parameters. An initial prior for $\pi$ would be $\pi\sim Dirichlet(\alpha_1,...,\alpha_{100})$. Of course $\forall i:\alpha_i>0, \pi_i\ge0, \sum_i\pi_i=1$. As you have a uniform assumption regarding the identities in the keystroke, you might want to take here $\alpha_1=...=\alpha_{100}$ and give it a high value (which reflects your high certainty in the Uniform manner).

Next the expectation vector with prior $\mu_i\sim U[1,10^6]$. Using the Bayes theorem we get $P(\mu|D)\propto P(D|\mu)P(\mu)$ (the denominator is a normalization constant which we can overlook for the moment). As the likelihood $P(D|\mu)$ is Gaussian and the prior $P(\mu)$ is uniform, the posterior $P(\mu|D)$ is a truncated Gaussian.

We start by choosing initial values for $\pi_i,\mu_i$ (for all $i$ values) and denote them $\pi_i^{(0)},\mu_i^{(0)}$. We then work iteratively, at each iteration $t+1$ we calculate the following steps, each step is calculated for all $k$ samples:

1. Responsibility $q_{ji}^{(t+1)}$ is the predicted probability of the sample $D_j$ belonging to the $i^{th}$ cluster after step $t$: $$q_{ji}^{(t+1)}=\frac{\pi_i^{(t)}\mathcal{N}(D_j;\mu_i^{(t)},1)}{\sum_{i=1}^N \pi_i^{(t)}\mathcal{N}(D_j;\mu_i^{(t)},1)}$$

2. Affiliation $$k_j^{(t+1)}\sim Multinomial \left(q_{j1}^{(t+1)},q_{j2}^{(t+1)},...,q_{jN}^{(t+1)}\right)$$

3. Cluster sizes: $$N_i^{(t+1)}=\sum_{j=1}^k\mathbb{I}\{k_j^{(t+1)}=j\}$$

4. Sampling the probabilities vector: $$(\pi^{(t+1)}|k^{(t+1)},D)\sim Dirichlet\left(N_1^{(t+1)}+\alpha_1,...,N_N^{(t+1)}+\alpha_N\right)$$

5. Calculating the expectations vector: Let $D^{i,(t+1)}$ be the subset of samples belonging to the $i^{th}$ cluster at this step: $$(\mu_i^{(t+1)}|k^{(t+1)},D^{i,(t+1)})=\frac{1}{N_i^{(t+1)}}\sum D^{i,(t+1)}$$

What happens eventually is that if the actual number of clusters is smaller that what we've initialized to, we'll get those clusters with 0 members and close to 0 mixture probability. You can also verify the number of clusters using the evidence function. Upon receiving a new sample $k+1$ you can re-run the whole process or (with some risks) initialize to the previous values.

Note that for the whole process, we have sticked with the basic assumptions and took nothing more.

We can predictively manipulate our inference regardless of what data we will get. If I decide to bug the person in the room one more time, I know beforehand that my best estimate would be N=k+1 . So I don't even need to ask the person anymore. I can already update my belief!

As Cliff noted, this isn't necessarily true. Assuming $k<N$ samples with $k$ different results $d_1,...,d_k$, you're always equally likely to get $d_{k+1}$ very close to one of the previous $k$ samples. This is true even for a certain range of $k>N$ values, as there's no guarantee that upon hitting $k=N$ requests you have been "exposed" to all $N$ options. The only thing that is guaranteed here is that for $K\ge N+1$ you have at least one cluster with two or more samples, due to the pigeonhole principle.

Taking no assumptions other than $N\le100$, you can solve this with Bayesian GMM and the Gibbs sampling algorithm. As you have known variance for the Gaussian models themselves and a strong assumption regarding the mixture probabilities. However we still take the probabilities vector $\pi$ as unknown so we don't limit ourselves to a fixed number of clusters (you'll see later why this works).

We denote $\theta=\{\pi,\{\mu_i\}_{i=1}^100\}$ our vector of parameters. An initial prior for $\pi$ would be $\pi\sim Dirichlet(\alpha_1,...,\alpha_{100})$. Of course $\forall i:\alpha_i>0, \pi_i\ge0, \sum_i\pi_i=1$. As you have a uniform assumption regarding the identities in the keystroke, you might want to take here $\alpha_1=...=\alpha_{100}$ and give it a high value (which reflects your high certainty in the Uniform manner).

Next the expectation vector with prior $\mu_i\sim U[1,10^6]$. Using the Bayes theorem we get $P(\mu|D)\propto P(D|\mu)P(\mu)$ (the denominator is a normalization constant which we can overlook for the moment). As the likelihood $P(D|\mu)$ is Gaussian and the prior $P(\mu)$ is uniform, the posterior $P(\mu|D)$ is a truncated Gaussian.

We start by choosing initial values for $\pi_i,\mu_i$ (for all $i$ values) and denote them $\pi_i^{(0)},\mu_i^{(0)}$. We then work iteratively, at each iteration $t+1$ we calculate the following steps, each step is calculated for all $k$ samples:

1. Responsibility $q_{ji}^{(t+1)}$ is the predicted probability of the sample $D_j$ belonging to the $i^{th}$ cluster after step $t$: $$q_{ji}^{(t+1)}=\frac{\pi_i^{(t)}\mathcal{N}(D_j;\mu_i^{(t)},1)}{\sum_{i=1}^N \pi_i^{(t)}\mathcal{N}(D_j;\mu_i^{(t)},1)}$$

2. Affiliation $$k_j^{(t+1)}\sim Multinomial \left(q_{j1}^{(t+1)},q_{j2}^{(t+1)},...,q_{jN}^{(t+1)}\right)$$

3. Cluster sizes: $$N_i^{(t+1)}=\sum_{j=1}^k\mathbb{I}\{k_j^{(t+1)}=j\}$$

4. Sampling the probabilities vector: $$(\pi^{(t+1)}|k^{(t+1)},D)\sim Dirichlet\left(N_1^{(t+1)}+\alpha_1,...,N_N^{(t+1)}+\alpha_N\right)$$

5. Caltulating the expectations vector: Let $D^{i,(t+1)}$ be the subset of samples belonging to the $i^{th}$ cluster at this step: $$(\mu_i^{(t+1)}|k^{(t+1)},D^{j,(t+1)})=\frac{1}{N_i^{(t+1)}}\sum D^{i,(t+1)}$$

What happens eventually is that if the actual number of clusters is smaller that what we've initialized to, we'll get those clusters with 0 members and close to 0 mixture probability. You can also verify the number of clusters using the evidence function. Upon receiving a new sample $k+1$ you can re-run the whole process or (with some risks) initialize to the previous values.

Note that for the whole process, we have sticked with the basic assumptions and took nothing more.

We can predictively manipulate our inference regardless of what data we will get. If I decide to bug the person in the room one more time, I know beforehand that my best estimate would be N=k+1 . So I don't even need to ask the person anymore. I can already update my belief!

As Cliff noted, this isn't necessarily true. Assuming $k<N$ samples with $k$ different results $d_1,...,d_k$, you're always equally likely to get $d_{k+1}$ very close to one of the previous $k$ samples. This is true even for a certain range of $k>N$ values, as there's no guarantee that upon hitting $k=N$ requests you have been "exposed" to all $N$ options. The only thing that is guaranteed here is that for $K\ge N+1$ you have at least one cluster with two or more samples, due to the pigeonhole principle.

Taking no assumptions other than $N\le100$, you can solve this with Bayesian GMM and the Gibbs sampling algorithm. As you have known variance for the Gaussian models themselves and a strong assumption regarding the mixture probabilities. However we still take the probabilities vector $\pi$ as unknown so we don't limit ourselves to a fixed number of clusters (you'll see later why this works).

We denote $\theta=\{\pi,\{\mu_i\}_{i=1}^100\}$ our vector of parameters. An initial prior for $\pi$ would be $\pi\sim Dirichlet(\alpha_1,...,\alpha_{100})$. Of course $\forall i:\alpha_i>0, \pi_i\ge0, \sum_i\pi_i=1$. As you have a uniform assumption regarding the identities in the keystroke, you might want to take here $\alpha_1=...=\alpha_{100}$ and give it a high value (which reflects your high certainty in the Uniform manner).

Next the expectation vector with prior $\mu_i\sim U[1,10^6]$. Using the Bayes theorem we get $P(\mu|D)\propto P(D|\mu)P(\mu)$ (the denominator is a normalization constant which we can overlook for the moment). As the likelihood $P(D|\mu)$ is Gaussian and the prior $P(\mu)$ is uniform, the posterior $P(\mu|D)$ is a truncated Gaussian.

We start by choosing initial values for $\pi_i,\mu_i$ (for all $i$ values) and denote them $\pi_i^{(0)},\mu_i^{(0)}$. We then work iteratively, at each iteration $t+1$ we calculate the following steps, each step is calculated for all $k$ samples:

1. Responsibility $q_{ji}^{(t+1)}$ is the predicted probability of the sample $D_j$ belonging to the $i^{th}$ cluster after step $t$: $$q_{ji}^{(t+1)}=\frac{\pi_i^{(t)}\mathcal{N}(D_j;\mu_i^{(t)},1)}{\sum_{i=1}^N \pi_i^{(t)}\mathcal{N}(D_j;\mu_i^{(t)},1)}$$

2. Affiliation $$k_j^{(t+1)}\sim Multinomial \left(q_{j1}^{(t+1)},q_{j2}^{(t+1)},...,q_{jN}^{(t+1)}\right)$$

3. Cluster sizes: $$N_i^{(t+1)}=\sum_{j=1}^k\mathbb{I}\{k_j^{(t+1)}=j\}$$

4. Sampling the probabilities vector: $$(\pi^{(t+1)}|k^{(t+1)},D)\sim Dirichlet\left(N_1^{(t+1)}+\alpha_1,...,N_N^{(t+1)}+\alpha_N\right)$$

5. Calculating the expectations vector: Let $D^{i,(t+1)}$ be the subset of samples belonging to the $i^{th}$ cluster at this step: $$(\mu_i^{(t+1)}|k^{(t+1)},D^{i,(t+1)})=\frac{1}{N_i^{(t+1)}}\sum D^{i,(t+1)}$$

What happens eventually is that if the actual number of clusters is smaller that what we've initialized to, we'll get those clusters with 0 members and close to 0 mixture probability. You can also verify the number of clusters using the evidence function. Upon receiving a new sample $k+1$ you can re-run the whole process or (with some risks) initialize to the previous values.

Note that for the whole process, we have sticked with the basic assumptions and took nothing more.