Two Independent variables have Bernoulli distribution: $X_1$ with $b(n,p)$ and $X_2$ with $b(m,p)$. How can I find conditional distribution $\mathbb P(X_1X_1+X_2=t)$?

If rv $X$ has [Bernoulli distribution](https://en.wikipedia.org/wiki/Bernoulli_distribution) then it only takes values in $\{0,1\}$. Don't you mean "binomial distribution"? If not then what is the role of parameters $n$ and $m$? Also I suspect you used the wrong tag "distributiontheory". It should be "probabilitydistributions". – drhab Apr 13 '19 at 13:42
3 Answers
Guide (preassuming that we are dealing with binomial distribution, see my comment on the question):
$$P(X_1=k\mid X_1+X_2=t)P(X_1+X_2=t)=P(X_1=k, X_2=tk)=P(X_1=k)P(X_2=tk)$$where the second equality rests on independence.
Further $X_1+X_2$ will have binomial distribution with parameters $n+m$ and $p$ (this also rests on independence of $X_1$ and $X_2$ and for a proof of this see here), enabling you to find and expression for $P(X_1+X_2=t)$.
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Hint: I assume that $X_1$ and $X_2$ are binomial distributed. We have $X_2=tX_1$. Now you can apply the Bayes theorem. Due independency of $X_1$ and $X_2$ we get
$$P(X_1=x_1X_1+X_2=t)=\frac{P(X_1=x_1)\cdot P(X_2=tx_1)}{P(X_1+X_2=t)}$$
where $X_1+X_2\sim Bin(n+m,p)$
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Assuming you meant Binomial random variables and that they are independents, you can always write $X_1=\sum_{i=1}^n Y_i$ and $X_2=\sum_{i=n+1}^{n+m} Y_i$ where the random variables are independent Bernoulli random variable with parameter $p$. Once you have that then $X_1+X_2=\sum_{i=1}^{n+m} Y_i$. The event $X_1+X_2=t$ just indicate that $t$ of the $Y_i$ are equal to $1$, conditioned on that the event $X_1=s$ is then an indication that $s$ of the first $n$ $Y_i$ are one. Observe that this implies that $ts$ of the last $m$ $Y_i$ are one, hence the probability is just the number of way of choosing $s$ Bernoulli out of the $n$ first and $ts$ Bernoulli out of the last $m$, we divide this by the amount of ways of having $t$ of the Bernoulli equal to $1$, hence \begin{align*} \mathbb P(X_1=sX_1+X_2=t) = \frac{{n\choose s}{m\choose ts}}{{m+n\choose t}} \end{align*}
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