$X_i\sim \text{Binom}(n_i,p_i)$, $X\sim \text{Binom}(n,p)$ and $X_1+X_2+...+X_k = X$ in distribution then show $\sum n_i = n$ and $p_i=p$

Question

$X_j\sim \operatorname{Binom}(n_j,p_j)$, $X\sim \operatorname{Binom}(n,p)$ and $X_1+X_2+...+X_k = X$ in distribution then show $\sum n_j = n$ and $p_j=p$

Using Characteristic Function I can conclude that $$(1-p+pe^{it})^n=\prod_{j=1}^k(1-p_j+p_je^{it})^{n_j}$$

I tried to put various values of $t(\in\mathbb R)$ to no avail. I know $\operatorname{Binom}(n,p)+\operatorname{Binom}(m,p)=\operatorname{Binom}(n+m,p)$ in distribution. How to proceed?

Answer 1

Incomplete answer (too much for a comment)

Proof of $\sum_{j=1}^kn_j=n$ under extra condition $p>0$.

---

Denote $r_m:=P(X=m)$ for every nonnegative integers $m$ and denote $Y=X_1+\cdots+X_k$.

If $p>0$ then $r_n>0$ and $r_m=0$ for every $m>n$.

Then $P(Y=n)=P(X=n)=r_n>0$.

From this it can be deduced that $n_1+\cdots+n_k\geq n$.

Also $P(Y=m)=P(X=m)=r_m=0$ for every $m>n$.

From this it can be deduced that $n_1+\cdots+n_k\leq n$.