Mean and variance of a probability distribution of increasing binomial coefficients

Question

I have this discrete random variable with probability mass function as $P(X=k) = aC_{n+k}^kp^k$ which I want to calculate the mean and variance.

For the mean what I tried is this:\begin{align*} E(X)&=a\sum_{k=0}^\infty kC_{n+k}^kp^k=ap\sum_{k=0}^\infty kC_{n+k}^kp^{k-1}\\ &=ap\sum_{k=0}^\infty C_{n+k}^k\frac{\mathrm d}{\mathrm dp}(p^k)=ap\frac{\mathrm d}{\mathrm dp}\left(\sum_{k=0}^\infty C_{n+k}^k p^k\right)\end{align*} Then I used a formula I found on the internet which says that $\sum\limits_{k=0}^\infty C_{n+k}^k p^k = \dfrac{p^n}{(1-p)^{n+1}}$ and I found $ E(X) = \dfrac{ap^n(n+p)}{(1-p)^{n+2}}$.

Then for the variance:\begin{align*} Var(X) &= a \sum_{k=0}^\infty k^2 C_{n+k}^k p^k - (E(X))^{2}\\ &= a \sum_{k=0}^\infty (k^2-k+k)C_{n+k}^k p^k - (E(X))^{2}\\ &= a \sum_{k=0}^\infty k(k-1)C_{n+k}^k p^k + E(X) - (E(X))^{2}\\ &= ap^2 \sum_{k=0}^\infty k(k-1)C_{n+k}^k p^{k-2} + E(X) - (E(X))^{2}\\ &= ap^2 \frac{\mathrm d^2}{\mathrm dp^2}\left(\sum_{k=0}^\infty C_{n+k}^k p^k) + E(X) - (E(X)\right)^{2}\end{align*}

I don't know if I went right and for the variance I can't simplify it pretty well. Is there a simpler method for doing this??

Thanks in advance.

Answer 1

$\def\F{\mathscr{F}}\def\peq{\mathrel{\phantom{=}}{}}$First, the identity $\sum\limits_{k = 0}^∞ \dbinom{n + k}{k} p^k = \dfrac{p^n}{(1 - p)^{n + 1}}$ is incorrect and it should be$$ \sum_{k = 0}^∞ \binom{n + k}{k} p^k = \frac{1}{(1 - p)^{n + 1}} $$ by differentiating $n$ times with respect to $p$ on both sides of $\sum\limits_{k = 0}^∞ p^k = \dfrac{1}{1 - p}$. The above identity also implies that $a = (1 - p)^{n + 1}$.

Because\begin{align*} E(X) &= \sum_{k = 0}^∞ k · \binom{n + k}{k} (1 - p)^{n + 1} p^k = \sum_{k = 1}^∞ k · \frac{(n + k)!}{k!\, n!} (1 - p)^{n + 1} p^k\\ &= \frac{p}{1 - p} \sum_{k = 1}^∞ (n + 1) · \frac{(n + k)!}{(k - 1)!\, (n + 1)!} (1 - p)^{n + 2} p^{k - 1}\\ &= \frac{p}{1 - p}(n + 1) \sum_{k = 0}^∞ \frac{(n + 1 + k)!}{k!\, (n + 1)!} (1 - p)^{n + 2} p^k = \frac{p}{1 - p}(n + 1), \end{align*}\begin{align*} E(X(X - 1)) &= \sum_{k = 0}^∞ k(k - 1) · \binom{n + k}{k} (1 - p)^{n + 1} p^k\\ &= \sum_{k = 2}^∞ k(k - 1) · \frac{(n + k)!}{k!\, n!} (1 - p)^{n + 1} p^k\\ &= \frac{p^2}{(1 - p)^2} \sum_{k = 2}^∞ (n + 1)(n + 2) · \frac{(n + k)!}{(k - 2)!\, (n + 2)!} (1 - p)^{n + 3} p^{k - 2}\\ &= \frac{p^2}{(1 - p)^2} (n + 1)(n + 2) \sum_{k = 0}^∞ \frac{(n + 2 + k)!}{k!\, (n + 2)!} (1 - p)^{n + 3} p^k\\ &= \frac{p^2}{(1 - p)^2} (n + 1)(n + 2), \end{align*} so $E(X) = \dfrac{p}{1 - p}(n + 1)$ and\begin{align*} D(X) &= E(X(X - 1)) + E(X) - (E(X))^2\\ &= \frac{p^2}{(1 - p)^2} (n + 1)(n + 2) + \frac{p}{1 - p}(n + 1) - \left( \frac{p}{1 - p}(n + 1) \right)^2\\ &= \frac{p}{(1 - p)^2} (n + 1). \end{align*}

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There is also a probabilistic approach using martingales. Suppose $Y_1, Y_2, \cdots$ are i.i.d. random variables on the probability space $({\mit Ω}, P, \F)$ with $P(Y_k = 0) = p$ and $P(Y_k = 1) = 1 - p$. Define$$ \F_m = σ(Y_1, \cdots, Y_m),\ S_m = \sum_{k = 1}^m Y_k,\ τ_m = \inf\{k \geqslant 0 \mid S_k = m\}, \quad \forall m \geqslant 0 $$ then $τ_m$ is a stopping time with respect to $\{\F_k\}_{k \geqslant 0}$ and $X \stackrel{\mathrm{d}}{=} τ_{n + 1} - (n + 1)$. So it remains to compute $E(τ_{n + 1})$ and $E(τ_{n+ 1}^2)$.

For $m \geqslant 0$, because $S_m$ is $\F_m$-measurable, and $σ(Y_{m + 1})$ and $\F_m$ are independent, then\begin{gather*} E(S_{m + 1} \mid \F_m) = E(S_m \mid \F_m) + E(Y_{m + 1} \mid \F_m)\\ = S_m + E(Y_{m + 1}) = S_m + (1 - p),\\ E(S_{m + 1}^2 \mid \F_m) = E(S_m^2 \mid \F_m) + 2 E(S_m Y_{m + 1} \mid \F_m) + E(Y_{m + 1}^2 \mid \F_m)\\ = S_m^2 + 2 E(Y_{m + 1}) S_m + E(Y_{m + 1}^2) = S_m^2 + 2(1 - p) S_m + (1 - p), \end{gather*} which implies that$$ E(S_{m + 1} - (1 - p)(m + 1) \mid\F_m) = S_m - (1 - p)m, $$\begin{align*} &\peq E\left( S_{m + 1}^2 - 2(1 - p)(m + 1) S_{m + 1} + ((1 - p)^2 (m + 1)^2 - p(1 - p) (m + 1)) \,\middle|\, \F_m \right)\\ &= \left( S_m^2 + 2(1 - p) S_m + (1 - p) \right) - 2(1 - p)(m + 1) \left( S_m + (1 - p)\right)\\ &\peq + ((1 - p)^2 (m + 1)^2 - p(1 - p) (m + 1))\\ &= S_m^2 - 2(1 - p)m S_m + ((1 - p)^2 m^2 - p(1 - p) m). \end{align*} Therefore, $\{S_m - (1 - p)m\}_{m \geqslant 0}$ and $\{S_m^2 - 2(1 - p)m S_m + ((1 - p)^2 m^2 - p(1 - p) m)\}_{m \geqslant 0}$ are martingales.

For any $N \geqslant 0$, define $σ_N = τ_{n + 1} ∧ N$, then the optional stopping theorem shows that\begin{gather*} E(S_{σ_N} - (1 - p)σ_N) = 0,\\ E(S_{σ_N}^2 - 2(1 - p)σ_N S_{σ_N} + ((1 - p)^2 σ_N^2 - p(1 - p) σ_N)) = 0, \end{gather*} thus\begin{gather*} (1 - p) E(σ_N) = E(S_{σ_N}),\tag{1}\\ (1 - p)^2 E(σ_N^2) = -E(S_{σ_N}^2) + 2(1 - p) E(σ_N S_{σ_N}) + p(1 - p) E(σ_N).\tag{2} \end{gather*} Now, note that $S_{τ_{n + 1}} = n + 1$ on $\{τ_{n + 1} < ∞\}$ and $P(τ_{n + 1} < ∞) = 1$. Since $σ_N ↗ τ_{n + 1}$ as $N → ∞$, then making $N → ∞$ in (1) yields $(1 - p) E(τ_{n + 1}) = E(S_{τ_{n + 1}}) = n + 1$, or $E(τ_{n + 1}) = \dfrac{n + 1}{1 - p}$. And analogously making $N → ∞$ in (2) yields\begin{align*} (1 - p)^2 E(τ_{n + 1}^2) &= -E(S_{τ_{n + 1}}^2) + 2(1 - p) E(τ_{n + 1} S_{τ_{n + 1}}) + p(1 - p) E(τ_{n + 1})\\ &= -(n + 1)^2 + 2(1 - p) E((n + 1) τ_{n + 1}) + p(1 - p) E(τ_{n + 1})\\ &= (n + 1)^2 + p(n + 1). \end{align*} Thus\begin{gather*} E(X) = E(τ_{n + 1}) - (n + 1) = \frac{p}{1 - p} (n + 1),\\ D(X) = D(τ_{n + 1}) = E(τ_{n + 1}^2) - (E(τ_{n + 1}))^2\\ = \frac{1}{(1 - p)^2} ((n + 1)^2 + p(n + 1)) - \left( \frac{n + 1}{1 - p}\right)^2 = \frac{p}{(1 - p)^2} (n + 1). \end{gather*}

Answer 2

I'm not sure whether it's simpler, but you can unify your calculations using characteristic function.

Let's define $ \phi(\lambda) := E[e^{i\lambda X}]$. Then we have $$ \phi(\lambda) = a \sum\limits_{k=0}^{\infty} e^{i \lambda k} C^k_{n+k} p^k $$ Here we introduce dummy variable $s_{\lambda} = e^{i \lambda} p$. And $\phi(\lambda)$ in terms of this variable looks like $$ \phi(\lambda) = a \sum\limits_{k=0}^{\infty} C^k_{n+k} s_{\lambda}^k. $$ Then using your formula $ {\sum_{k=0}^\infty}C_{n+k}^k p^k = \frac{p^n}{(1-p)^{n+1}}$ we have $$ \phi(\lambda) = a \frac{s^n}{(1-s)^{n+1}} = a \frac{e^{i\lambda n} p^n}{(1-e^{i\lambda} p)^{n+1}} $$

Then you have the following formula for moments of arbitrary power in terms of charateristic function: $$ \phi^{(k)}(0) = i^k E[X^k] $$ Proof is pretty straightforward: $$ \phi^{(k)}(0) = \frac{d^k}{d \lambda^k} \phi(\lambda) \bigg\vert_{\lambda = 0} = E[i^k X^k e^{i\lambda X}]\bigg\vert_{\lambda = 0} = i^k E[X^k] $$

Now we can compute $E[X], \, E[X^2]$: $$ E[X] = -i \phi'(\lambda) = a\frac{p^n(p+n)}{(1-p)^{n+2}} $$ $$ E[X^2] = - \phi''(\lambda) = a \frac{p^n (n^2 + 3 n p + p^2 + p)}{(1 - p)^{n+3} } $$ And we have the same result.

Actually, I have done the same calculation as you, but written them another way. I find it neater. Hope it will help.