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How can I formally prove that the sum of two independent binomial variables X and Y with same parameter p is also a binomial ?

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    $\begingroup$ why was that question down voted? $\endgroup$ Commented Oct 25, 2015 at 3:49

4 Answers 4

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Let $(B_k)_k$ be a sequence of iid Bernoulli distributed random variable with $P(B_k=1)=p$ for $k=1,2,\dots$

Then $$X:=B_1+\cdots+B_n$$ is binomially distributed with parameters $n,p$ and $$Y:=B_{n+1}+\cdots+B_{n+m}$$ is binomially distributed with parameters $m,p$. It is evident that $X$ and $Y$ are independent.

Now realize that $$X+Y=B_1+\cdots+B_{n+m}$$ is binomially distributed with parameters $n+m,p$.

This spares you any computations.

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    $\begingroup$ Thanks for your answer. Can you explain in some more details how did you get to your first result ? $\endgroup$ Commented Mar 9, 2015 at 20:54
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    $\begingroup$ If $X$ has binomial distribution with parameters $n,p$ then it can be identified as the number of "successes" by having $n$ independent experiments all having a probability of $p$ to succeed. Set $B_i=1$ if experiment $i$ succeeds and $B_i=0$ otherwise. This for $i=1,\dots,n$. Then the number of successes (that is $X$) equals $B_1+\cdots +B_n$. $\endgroup$
    – drhab
    Commented Mar 10, 2015 at 9:44
  • $\begingroup$ Oh sorry I misread the term Bernouilli as Binomial. Thank you so much, this is very easy to reason about. $\endgroup$ Commented Mar 10, 2015 at 17:35
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Just compute. Suppose $X \sim \def\Bin{\mathord{\rm Bin}}\Bin(n,p)$, $Y \sim \Bin(m,p)$. Now let $0 \le k \le n+m$, then \begin{align*} \def\P{\mathbb P}\P(X+Y = k) &= \sum_{i=0}^k \P(X = i, Y = k-i)\\ &= \sum_{i=0}^k \P(X=i)\P(Y=k-i) & \text{by independence}\\ &= \sum_{i=0}^k \binom ni p^i (1-p)^{n-i} \binom m{k-i} p^{k-i} (1-p)^{m-k+i}\\ &= p^k(1-p)^{n+m-k}\sum_{i=0}^k \binom ni \binom m{k-i} \\ &= \binom {n+m}k p^k (1-p)^{n+m-k} \end{align*} Hence $X+Y \sim \Bin(n+m,p)$.

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    $\begingroup$ Thanks for your answer. Can you please explain how is $ \sum_{i=0}^k \binom ni \binom m{k-i} $ equal to $ \binom {n+m}{k} $ ? $\endgroup$ Commented Mar 5, 2015 at 19:33
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    $\begingroup$ @Piyush see Vandermonde's identity $\endgroup$
    – polpetti
    Commented Jan 3, 2017 at 20:13
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Another way: Suppose $X\sim$ Bin$(n, p)$ and $Y\sim$ Bin$(m, p)$. The characteristic function of $X$ is then $$\varphi_X(t) = E[e^{itX}]=\sum_{k=0}^ne^{itk}{n\choose k}p^k(1-p)^{n-k}=\sum_{k=0}^n{n\choose k} (pe^{it})^k(1-p)^{n-k}=(1-p+pe^{it})^n.$$

Since $X, Y$ independent, $$\varphi_{X+Y}(t)=\varphi_{X}(t)\varphi_Y(t)=(1-p+pe^{it})^n(1-p+pe^{it})^m=(1-p+pe^{it})^{n+m}.$$

By uniqueness, we get $X+Y\sim$ Bin$(n+m, p)$.

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We can prove this using Moment generating function as follows if someone is not comfortable with characterstic functions as answered above:

Let $X \sim B(n_1,p_1)$ and $Y \sim B(n_2,p_2)$ be independent random variables.

We know the MGF of the Binomial distribution is as follows:

$M_X(t)=(q_1+p_1e^t)^{n_{1}},M_Y(t)=(q_2+p_2e^t)^{n_{2}} $

Since X and Y are independent

$M_{X+Y}(t)=M_X(t) \cdot M_y(t) =(q_1+p_1e^t)^{n_{1}} \cdot (q_2+p_2e^t)^{n_{2}}$

We see that we cannot express it in the form $(q+pe^t)^{n}$ and thus by uniqueness property of MGF $X+Y$ is not a binomial variate. However if we take $p_1=p_2=p$ then we have:

$M_{X+Y}(t)=M_X(t) \cdot M_y(t) =(q+pe^t)^{n_{1}} \cdot (q+pe^t)^{n_{2}}$ $=(q+pe^t)^{n_{1}+n_{2}} $

which is MGF of binomial variate with parameters $(n_1+n_2,p)$

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