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Based on the comments to this answer, I no longer believe what I initially wrote (still appearing at the bottom of the answer). It seems to me a construction should be possible. It is at least possible in the case of a Binomial point process.

Let $\{X_i\}_{i \in \mathbb{N},i\neq 4}$ be independent Bernoulli$(1/2)$ random variables. Let $X_4'$ be Bernoulli$(1/2)$ and independent of the $X_i$. Then define $X_4$ as follows:

• $X_4 = 1$ if $(X_1,X_2,X_3)$ is either $(0,0,1)$ or $(1,1,0)$,
• $X_3 = 0$ if $(X_1,X_2,X_3)$ is either $(0,1,0)$ or $(1,0,1)$,
• $X_4=X_4'$ otherwise.

Then for all $j \geq 0$, $n \geq 1$, $X_{j+1}+\ldots+X_{j+n}$ has Binomial$(n,1/2)$ distribution but the family $(X_n)_{n \in \mathbb{N}}$ are not iid.

What appears below (where I suggested such a construction was impossible) is false.

For a standard Poisson process, this won't be possible. (See this question and its answer.)

Edit: Given the comments perhaps I should provide more detail.

With probability one, for every pair $0 < p < q$, $p,q$ rationals, $X(q)−X(p)$ is a non-negative integer. Since X is cadlag the same property must hold for every real pair $0 < s < t$, i.e. $X$ is increasing and integer-valued.

Let us also show that $X$ has no jumps of size more than one: with probability one, for all $x > 0$, $X(x^-) := \lim_{y \uparrow x} X(y) \geq X(x)-1$. If this failed to hold then there would be $\epsilon > 0$ and $t < \infty$ so that $$ \mathbb{P}(\exists x \in [0,t), X(x)-X(x^-) \geq 2) > \epsilon/2. $$ But since $X$ is increasing, for any positive integer $n$ we can bound this probability from above by $$ \sum_{1 \leq i < 2n} \mathbb{P}(X((i+1)t/2n)-X((i-1)t/2n) \geq 2) $$ the point being that these intervals are chosen to overlap so that a jump of size $\geq 2$ must fall in at least one of them. Each of the differences above is distributed as Poisson$(t/n)$, so the associated probability is $o(n^{-1})$ as $n \to \infty$ and thus the whole sum tends to zero as $n \to \infty$.

We then know that a process $X$ such as you describe must be increasing and integer valued, with all jumps of size $1$. In other words, $X$ is a point process on $[0,\infty)$. Now the answer from the other thread implies that $X$ must be a rate one Poisson process.

Based on the comments to this answer, I no longer believe what I initially wrote (still appearing at the bottom of the answer). It seems to me a construction should be possible. It is at least possible in the case of a Binomial point process.

Let $\{X_i\}_{i \in \mathbb{N},i\neq 4}$ be independent Bernoulli$(1/2)$ random variables. Let $X_4'$ be Bernoulli$(1/2)$ and independent of the $X_i$. Then define $X_4$ as follows:

• $X_4 = 1$ if $(X_1,X_2,X_3)$ is either $(0,0,1)$ or $(1,1,0)$,
• $X_3 = 0$ if $(X_1,X_2,X_3)$ is either $(0,1,0)$ or $(1,0,1)$,
• $X_4=X_4'$ otherwise.

Then for all $j \geq 0$, $n \geq 1$, $X_{j+1}+\ldots+X_{j+n}$ has Binomial$(n,1/2)$ distribution but the family $(X_n)_{n \in \mathbb{N}}$ are not iid.

What appears below (where I suggested such a construction was impossible) is false.

For a standard Poisson process, this won't be possible. (See this question and its answer.)

Edit: Given the comments perhaps I should provide more detail.

With probability one, for every pair $0 < p < q$, $p,q$ rationals, $X(q)−X(p)$ is a non-negative integer. Since X is cadlag the same property must hold for every real pair $0 < s < t$, i.e. $X$ is increasing and integer-valued.

Let us also show that $X$ has no jumps of size more than one: with probability one, for all $x > 0$, $X(x^-) := \lim_{y \uparrow x} X(y) \geq X(x)-1$. If this failed to hold then there would be $\epsilon > 0$ and $t < \infty$ so that $$ \mathbb{P}(\exists x \in [0,t), X(x)-X(x^-) \geq 2) > \epsilon/2. $$ But since $X$ is increasing, for any positive integer $n$ we can bound this probability from above by $$ \sum_{1 \leq i < 2n} \mathbb{P}(X((i+1)t/2n)-X((i-1)t/2n) \geq 2) $$ the point being that these intervals are chosen to overlap so that a jump of size $\geq 2$ must fall in at least one of them. Each of the differences above is distributed as Poisson$(t/n)$, so the associated probability is $o(n^{-1})$ as $n \to \infty$ and thus the whole sum tends to zero as $n \to \infty$.

We then know that a process $X$ such as you describe must be increasing and integer valued, with all jumps of size $1$. In other words, $X$ is a point process on $[0,\infty)$. Now the answer from the other thread implies that $X$ must be a rate one Poisson process.

Based on the comments to this answer, I no longer believe what I initially wrote (still appearing at the bottom of the answer). It seems to me a construction should be possible. It is at least possible in the case of a Binomial point process.

Let $\{X_i\}_{i \in \mathbb{N},i\neq 3}$ be independent Bernoulli$(1/2)$ random variables. Let $X_3'$ be Bernoulli$(1/2)$ and independent of the $X_i$. Then let $X_3 = 1$ if $(X_1,X_2)=(1,0)$, $X_3=0$ if $(X_1,X_2)=(0,1)$, and $X_3=X_3'$ otherwise.

Then for all $j \geq 0$, $n \geq 1$, $X_{j+1}+\ldots+X_{j+n}$ has Binomial$(n,1/2)$ distribution but the family $(X_n)_{n \in \mathbb{N}}$ are not iid.

What appears below the line (where I suggested such a construction was impossible) is wrong, because I have no proof that a point process is determined by its distribution on intervals.

For a standard Poisson process, this won't be possible. (See this question and its answer.)

Edit: Given the comments perhaps I should provide more detail.

With probability one, for every pair $0 < p < q$, $p,q$ rationals, $X(q)−X(p)$ is a non-negative integer. Since X is cadlag the same property must hold for every real pair $0 < s < t$, i.e. $X$ is increasing and integer-valued.

Let us also show that $X$ has no jumps of size more than one: with probability one, for all $x > 0$, $X(x^-) := \lim_{y \uparrow x} X(y) \geq X(x)-1$. If this failed to hold then there would be $\epsilon > 0$ and $t < \infty$ so that $$ \mathbb{P}(\exists x \in [0,t), X(x)-X(x^-) \geq 2) > \epsilon/2. $$ But since $X$ is increasing, for any positive integer $n$ we can bound this probability from above by $$ \sum_{1 \leq i < 2n} \mathbb{P}(X((i+1)t/2n)-X((i-1)t/2n) \geq 2) $$ the point being that these intervals are chosen to overlap so that a jump of size $\geq 2$ must fall in at least one of them. Each of the differences above is distributed as Poisson$(t/n)$, so the associated probability is $o(n^{-1})$ as $n \to \infty$ and thus the whole sum tends to zero as $n \to \infty$.

We then know that a process $X$ such as you describe must be increasing and integer valued, with all jumps of size $1$. In other words, $X$ is a point process on $[0,\infty)$. Now the answer from the other thread implies that $X$ must be a rate one Poisson process.

Based on the comments to this answer, I no longer believe what I initially wrote (still appearing at the bottom of the answer). It seems to me a construction should be possible. It is at least possible in the case of a Binomial point process.

Let $\{X_i\}_{i \in \mathbb{N},i\neq 4}$ be independent Bernoulli$(1/2)$ random variables. Let $X_4'$ be Bernoulli$(1/2)$ and independent of the $X_i$. Then define $X_4$ as follows:

• $X_4 = 1$ if $(X_1,X_2,X_3)$ is either $(0,0,1)$ or $(1,1,0)$,
• $X_3 = 0$ if $(X_1,X_2,X_3)$ is either $(0,1,0)$ or $(1,0,1)$,
• $X_4=X_4'$ otherwise.

Then for all $j \geq 0$, $n \geq 1$, $X_{j+1}+\ldots+X_{j+n}$ has Binomial$(n,1/2)$ distribution but the family $(X_n)_{n \in \mathbb{N}}$ are not iid.

What appears below (where I suggested such a construction was impossible) is false.

For a standard Poisson process, this won't be possible. (See this question and its answer.)

Edit: Given the comments perhaps I should provide more detail.

With probability one, for every pair $0 < p < q$, $p,q$ rationals, $X(q)−X(p)$ is a non-negative integer. Since X is cadlag the same property must hold for every real pair $0 < s < t$, i.e. $X$ is increasing and integer-valued.

Let us also show that $X$ has no jumps of size more than one: with probability one, for all $x > 0$, $X(x^-) := \lim_{y \uparrow x} X(y) \geq X(x)-1$. If this failed to hold then there would be $\epsilon > 0$ and $t < \infty$ so that $$ \mathbb{P}(\exists x \in [0,t), X(x)-X(x^-) \geq 2) > \epsilon/2. $$ But since $X$ is increasing, for any positive integer $n$ we can bound this probability from above by $$ \sum_{1 \leq i < 2n} \mathbb{P}(X((i+1)t/2n)-X((i-1)t/2n) \geq 2) $$ the point being that these intervals are chosen to overlap so that a jump of size $\geq 2$ must fall in at least one of them. Each of the differences above is distributed as Poisson$(t/n)$, so the associated probability is $o(n^{-1})$ as $n \to \infty$ and thus the whole sum tends to zero as $n \to \infty$.

We then know that a process $X$ such as you describe must be increasing and integer valued, with all jumps of size $1$. In other words, $X$ is a point process on $[0,\infty)$. Now the answer from the other thread implies that $X$ must be a rate one Poisson process.

For a standard Poisson process, this won't be possible. (See this question and its answer.)

Edit: Given the comments perhaps I should provide more detail.

With probability one, for every pair $0 < p < q$, $p,q$ rationals, $X(q)−X(p)$ is a non-negative integer. Since X is cadlag the same property must hold for every real pair $0 < s < t$, i.e. $X$ is increasing and integer-valued.

Let us also show that $X$ has no jumps of size more than one: with probability one, for all $x > 0$, $X(x^-) := \lim_{y \uparrow x} X(y) \geq X(x)-1$. If this failed to hold then there would be $\epsilon > 0$ and $t < \infty$ so that $$ \mathbb{P}(\exists x \in [0,t), X(x)-X(x^-) \geq 2) > \epsilon/2. $$ But since $X$ is increasing, for any positive integer $n$ we can bound this probability from above by $$ \sum_{1 \leq i < 2n} \mathbb{P}(X((i+1)t/2n)-X((i-1)t/2n) \geq 2) $$ the point being that these intervals are chosen to overlap so that a jump of size $\geq 2$ must fall in at least one of them. Each of the differences above is distributed as Poisson$(t/n)$, so the associated probability is $o(n^{-1})$ as $n \to \infty$ and thus the whole sum tends to zero as $n \to \infty$.

We then know that a process $X$ such as you describe must be increasing and integer valued, with all jumps of size $1$. In other words, $X$ is a point process on $[0,\infty)$. Now the answer from the other thread implies that $X$ must be a rate one Poisson process.

Based on the comments to this answer, I no longer believe what I initially wrote (still appearing at the bottom of the answer). It seems to me a construction should be possible. It is at least possible in the case of a Binomial point process.

Let $\{X_i\}_{i \in \mathbb{N},i\neq 3}$ be independent Bernoulli$(1/2)$ random variables. Let $X_3'$ be Bernoulli$(1/2)$ and independent of the $X_i$. Then let $X_3 = 1$ if $(X_1,X_2)=(1,0)$, $X_3=0$ if $(X_1,X_2)=(0,1)$, and $X_3=X_3'$ otherwise.

Then for all $j \geq 0$, $n \geq 1$, $X_{j+1}+\ldots+X_{j+n}$ has Binomial$(n,1/2)$ distribution but the family $(X_n)_{n \in \mathbb{N}}$ are not iid.

What appears below the line (where I suggested such a construction was impossible) is wrong, because I have no proof that a point process is determined by its distribution on intervals.

For a standard Poisson process, this won't be possible. (See this question and its answer.)

Edit: Given the comments perhaps I should provide more detail.

With probability one, for every pair $0 < p < q$, $p,q$ rationals, $X(q)−X(p)$ is a non-negative integer. Since X is cadlag the same property must hold for every real pair $0 < s < t$, i.e. $X$ is increasing and integer-valued.

Let us also show that $X$ has no jumps of size more than one: with probability one, for all $x > 0$, $X(x^-) := \lim_{y \uparrow x} X(y) \geq X(x)-1$. If this failed to hold then there would be $\epsilon > 0$ and $t < \infty$ so that $$ \mathbb{P}(\exists x \in [0,t), X(x)-X(x^-) \geq 2) > \epsilon/2. $$ But since $X$ is increasing, for any positive integer $n$ we can bound this probability from above by $$ \sum_{1 \leq i < 2n} \mathbb{P}(X((i+1)t/2n)-X((i-1)t/2n) \geq 2) $$ the point being that these intervals are chosen to overlap so that a jump of size $\geq 2$ must fall in at least one of them. Each of the differences above is distributed as Poisson$(t/n)$, so the associated probability is $o(n^{-1})$ as $n \to \infty$ and thus the whole sum tends to zero as $n \to \infty$.

We then know that a process $X$ such as you describe must be increasing and integer valued, with all jumps of size $1$. In other words, $X$ is a point process on $[0,\infty)$. Now the answer from the other thread implies that $X$ must be a rate one Poisson process.