# How to compute the sum of geometric distribution [closed]

How to compute the sum of random variables of geometric distribution $X_{i}(i=0,1,2..n)$ is the independent random variables of geometric distribution, that is, $P(X_{i}=x)=p(1-p)^{x}$, then how to compute the PDF of $\sum_{i=1}^{n}X_{i}$

a negative binomial distribution $P(Y)=\binom{r+y-1}{y}p^{r}(1-p)^{y}$, Prove that sum of random variables of geometric distribution become a negative binomial distribution? how to do this deduction?? can help here? Thanks!!!

## closed as unclear what you're asking by Did, Najib Idrissi, Claude Leibovici, Shaun, TravisJan 24 '15 at 9:42

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• Possibly see what it's moment generating function of sum will correspond to – Kamster Jan 24 '15 at 7:40
• Thank you very much.You proof to me!!! – pan Jan 24 '15 at 7:58

A direct method would rely on induction: Let $Y \sim \operatorname{NegBinomial}(r,p)$ with $$\Pr[Y = y] = \binom{r+y-1}{y} p^r (1-p)^y, \quad y = 0, 1, 2, \ldots,$$ so $Y$ counts the random number of "failures" before the $r^{\rm th}$ "success" is observed, where the probability of "success" is $p$. Let $X \sim \operatorname{Geometric}(p)$ with $$\Pr[X = x] = p(1-p)^x, \quad x = 0, 1, \ldots,$$ which counts the number of "failures" before the first "success" is observed (i.e., this is a special case of the negative binomial for $r = 1$). Then consider the variable $W = Y+X$. By direct convolution, \begin{align*} \Pr[W = w] &= \Pr[X+Y = w] \\ &= \sum_{k=0}^w \Pr[(Y = k) \cap (X = w-k)] \\ &= \sum_{k=0}^w \binom{r+k-1}{k} p^r (1-p)^k \cdot p (1-p)^{w-k} \\ &= \sum_{k=0}^w \binom{r+k-1}{k} p^{r+1} (1-p)^w \\ &= p^{r+1} (1-p)^w \sum_{k=0}^w \binom{r+k-1}{k} \\ &= \binom{r+w}{w} p^{r+1} (1-p)^w, \end{align*} where the last equality follows from the "hockey stick" or "Christmas stocking" identity $$\sum_{i=0}^{k-1} \binom{n+i}{i} = \binom{n+k}{k-1}$$ with the choice $n = r-1$, $k = w+1$. This shows that $W$ is negative binomial with parameters $r^* = r+1$ and $p$, and it is now evident that if $X_i \sim \operatorname{Geometric}(p)$ for $i = 1, 2, \ldots, n$ are IID, then $\sum X_i \sim \operatorname{NegBinomial}(n,p)$.

Now for a different approach, let's use the probability generating function (PGF): $$P_X(t) = \operatorname{E}[t^X].$$ Let $S = \sum_{i=1}^n X_i$ where $X_i$ is defined as above. Then note that the PGF retains the property that the PGF of the sum of independent random variables equals the product of the PGFs of each random variable: $$P_S(t) = \operatorname{E}[t^S] = \operatorname{E}\left[t^{\sum X_i}\right] = \prod_{i=1}^n \operatorname{E}[t^{X_i}] = \prod_{i=1}^n P_{X_i}(t) = (P_X(t))^n.$$ So for a negative binomial PGF, we have \begin{align*} P_Y(t) &= \sum_{y=0}^\infty t^y \binom{r+y-1}{y} p^r (1-p)^y \\ &= \left(\frac{p}{1-(1-p)t}\right)^r \sum_{y=0}^\infty \binom{r+y-1}{y} (1-(1-p)t)^r ((1-p)t)^y \\ &= \left(\frac{p}{1-(1-p)t}\right)^r, \quad 0 < t < \frac{1}{1-p}.\end{align*} since the sum in the penultimate expression is simply the sum of a negative binomial PDF with parameters $r$ and $p^* = 1-(1-p)t$ over its support, and is therefore equal to $1$. From this calculation, we also get the PGF of a geometric distribution by setting $r = 1$: $$P_X(t) = \frac{p}{1-(1-p)t}, \quad 0 < t < \frac{1}{1-p}.$$ But now the relationship between the two is obvious: $$(P_X(t))^n = \left(\frac{p}{1-(1-p)t}\right)^n = P_S(t),$$ and the proof is complete by uniqueness of PGFs.

• Thank you very much. This really helps me a lot. – pan Jan 24 '15 at 8:38
• why consider the variable W=Y+X.thank you – pan Jan 26 '15 at 5:39
• The idea is to use induction: If $X_1, X_2, \ldots, X_{n-1}$ are all IID geometric, then assume that $Y = X_1 + X_2 + \cdots + X_{n-1}$ is negative binomial. Then if you can show $W = Y + X_n$ is also negative binomial, then that completes the induction step. – heropup Jan 26 '15 at 5:59

try computing the MGF of ∑Xi . the MGF of geometric Xi is p[(1-q(e^t))^(-1)] . where q=1=p. and t is a constant chosen such that the quantity converges. now MGF of ∑Xi will be the product such n quantities. (as Xi's are independent. ) and , consequently you will get the MGF to be p^n[(1-q(e^t))^(-n)]. which is MGF of negative binomial distribution. by uniqueness property of MGF , the result follows.

• specifically , t< ln(1/q) – saudade Jan 24 '15 at 7:50
• Thank you very much.You proof to me!!! – pan Jan 24 '15 at 7:58
• if you are happy with the answer, click on the 'tick' sign on the left hand side of my answer. doing that you 'accpet my answer' and 'change the status of your question as answered' . and for that you will get some points ( which will make you able to do more activity flexibly in the site ). thank you. – saudade Jan 24 '15 at 8:09