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The Problem: Let $X\thicksim\text{Binomial}(n,p)$ and $Y\thicksim\text{Bernoulli}(p)$ be independent random variables. Find the distribution of $X+Y.$

My Attempt: Set $Z=X+Y$. Then we can compute the distribution of $Z$ using the convolution formula because $X$ and $Y$ are independent. Also, since $X$ only takes values in $\{0,1,\dots,n\}$ and $Y$ only takes values in $\{0,1\}$, we have that $Z$ can only take values in $\{0,1,\dots,n+1\}$. Altogether, it follows that the probability mass function of $Z$ for $k\in\{0,1,\dots,n+1\}$ is given by \begin{equation*}\begin{split} p_Z(k)&=\sum_{\ell\in\mathbb Z}p_X(k-\ell)p_Y(\ell)=p_X(k)p_Y(0)+p_X(k-1)p_Y(1)\\ &=\binom{n}{k}p^k(1-p)^{n-k+1}+\binom{n}{k}p^{k+1}(1-p)^{n-k}. \end{split}\end{equation*}

My Concern: My issue with my work above is that we cannot plug in $n+1$ into $p_Z(k)$, which leads me to think that the range of $Z$ is $\{0,1,\dots,n\}$ instead. Could anyone please shed some light into this issue I am encountering? Also any comments on the rest of my attempt are very welcomed. Thank you very much for your time.

StubbornAtom
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Stackman
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1 Answers1

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You made a mistake in the second term.

Correct is: $$p_{Z}\left(k\right)=\binom{n}{k}p^{k}\left(1-p\right)^{n+1-k}+\binom{n}{k-1}p^{k}\left(1-p\right)^{n+1-k}=$$$$\left[\binom{n}{k}+\binom{n}{k-1}\right]p^{k}\left(1-p\right)^{n+1-k}=\binom{n+1}{k}p^{k}\left(1-p\right)^{n+1-k}$$

Also be aware that we can write $X:=B_1+\cdots+ B_n$ where the $B_i$ are iid with Bernoulli-distribution with parameter $p$.

This reveals more directly that $X+Y=B_1+\cdots+B_n+Y$ again has binomial distribution (this time with parameter $n+1$).

It might help to have a look at this answer.

drhab
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