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Proof that for Poisson independent processes with $0<s<t$,

$$P[x(s)=k | x(t)+y(t)=n]$$

has binomial distribution

$$\frac{n!}{k!(n-k)!}p^k(1-p)^{(n-k)}$$

With $p=\frac{\lambda_x}{\lambda_x+\lambda_y}\frac{s}{t}$


TIP

It is proved by Bayes and Total Probability Theorems that for $x$ and $y$ Poisson independent processes and $0<s<t$:

$$P[x(t)=k | x(t)+y(t)=n]$$

and

$$P[x(s)=k | x(t)=n]$$

have binomial distribution as having $k$ successes in $n$ tries with success provabilities respectively $p=\frac{\lambda_x}{\lambda_x+\lambda_y}$ and $\frac{s}{t}$.

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Both results you specified in "TIPS" follow from the observation that if $A$ and $B$ are independent Poisson r.v.s with parameters $a$ and $b$, respectively, the conditional distribution of $A$ given $A+B=n$ is binomial, with parameters $a/(a+b)$ and $n$. To see this, use the fact that $A+B$ is Poisson with parameter $a+b.$ Then the expression $$P(A=k|A+B=n)=\frac{P(A=k, A+B=n)}{P(A+B=n)} = \frac{P(A=k, B=n-k)}{P(A+B=n)}$$ evaluates to $$P(A=k|A+B=n)=\frac{P(A=a)P(B=n-k)}{P(A+B=n)}=\frac{e^{-a}a^k}{k!}\frac{e^{-b}b^{n-k}}{(n-k)!}\big/\frac{e^{a+b}(a+b)^n}{n!}$$ which simplifies to the stated binomial probability.

To apply this to the observations at hand, take $A=x(t)$ and $B=y(t)$ in the first instance, and $A=x(s)$ and $B=x(t)-x(s)$ in the second. Note that $x(t)-x(s)$ is independent from $x(s)$ because of the Poisson memoryless property and their parameters are $\lambda_xs$ and $\lambda_x(t-s)$.

Similarly you can prove your question by taking $A=x(s)$ and $B=x(t)+y(t)-x(s)$, so that the binomial probability success becomes:

$$\frac{a}{a+b}=\frac{\lambda_xs}{\lambda_xs+\lambda_x(t-s)+\lambda_yt}=\frac{\lambda_xs}{\lambda_xt+\lambda_yt}=\frac{\lambda_x}{\lambda_x+\lambda_y}\frac{s}{t}$$

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