Steven R. Dunbar
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Topics in
Probability Theory and Stochastic Processes
Steven R. Dunbar

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Stirling’s Formula derived from the Poisson Distribution

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### Rating

Mathematicians Only: prolonged scenes of intense rigor.

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### Section Starter Question

What is the Poisson distribution? What kind of circumstances does a Poisson distribution describe?

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### Key Concepts

1. Stirling’s Formula, also called Stirling’s Approximation, is the asymptotic relation
$n!\sim \sqrt{2\pi }{n}^{n+1∕2}{e}^{-n}.$

2. The formula is useful in estimating large factorial values, but its main mathematical value is in limits involving factorials.
3. The Poisson distribution with parameter $\lambda$ is the discrete probability distribution deﬁned on the non-negative integers $0,1,2,\dots$ with probability mass $ℙ\left[X=k\right]=p\left(k;\lambda \right)=\frac{{\lambda }^{k}}{k!}{e}^{-\lambda }$.

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### Vocabulary

1. Stirling’s Formula, also called Stirling’s Approximation, is the asymptotic relation
$n!\sim \sqrt{2\pi }{n}^{n+1∕2}{e}^{-n}.$

2. The Poisson distribution with parameter $\lambda$ is the discrete probability distribution deﬁned on the non-negative integers $0,1,2,\dots$ with probability mass $ℙ\left[X=k\right]=p\left(k;\lambda \right)=\frac{{\lambda }^{k}}{k!}{e}^{-\lambda }$.

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### Mathematical Ideas

#### Stirling’s Formula

Stirling’s Formula, also called Stirling’s Approximation, is the asymptotic relation

$n!\sim \sqrt{2\pi }{n}^{n+1∕2}{e}^{-n}.$

The formula is useful in estimating large factorial values, but its main mathematical value is in limits involving factorials. Another attractive form of Stirling’s Formula is

$n!\sim \sqrt{2\pi n}{\left(\frac{n}{e}\right)}^{n}.$

An improved inequality version of Stirling’s Formula is

$\sqrt{2\pi }{n}^{n+1∕2}{e}^{-n+1∕\left(12n+1\right)}

See Stirling’s Formula. in MathWorld.com.

#### Intuitive Probabilistic Proof

This intuitive argument is adapted from [6, page 171-172] and also from the short article by Hu, [4].

Let ${X}_{1},{X}_{2},\dots$ be independent Poisson random variables each having mean $1$. Let ${S}_{n}={\sum }_{j=1}^{n}{X}_{j}$ and then note that the mean and the variance of ${S}_{n}$ are equal to $n$.

$\begin{array}{llll}\hfill ℙ\left[{S}_{n}=n\right]& =ℙ\left[n-1<{S}_{n}\le n\right]\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =ℙ\left[-1∕\sqrt{n}<\left({S}_{n}-n\right)∕\sqrt{n}\le 0\right]\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx {\int }_{-1∕\sqrt{n}}^{0}\frac{1}{\sqrt{2\pi }}{e}^{-{x}^{2}∕2}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx \frac{1}{\sqrt{2\pi }}\frac{1}{\sqrt{n}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

On the other hand ${S}_{n}$ is Poisson with mean $n$, and so

$ℙ\left[{S}_{n}=n\right]=\frac{{e}^{-n}{n}^{n}}{n!}$

Equating the two expressions for $ℙ\left[{S}_{n}=n\right]$, and rearranging obtain

$n!\approx \sqrt{2\pi }{n}^{n+1∕2}{e}^{-n}.$

This “proof” relies on having a version of the Central Limit Theorem that has not been proved using Stirling’s Formula! Such a version of the Central Limit Theorem itself is pretty advanced. It also relies on several other advanced facts, such as the distribution of Poisson random variables, the fact that the sum of independent Poisson random variables is again Poisson and the fact that the variance of the sum of independent random variables is the sum of the variances. On the other hand, it is short and simple!

#### Rigorous Derivation of Stirling’s Formula

The characteristic function with variable $𝜃\in ℝ$ of the Poisson distribution $p\left(k;\lambda \right)=\frac{{\lambda }^{k}}{k!}{e}^{-\lambda }$ with parameter $\lambda$ is

$\stackrel{̂}{p}\left(𝜃;\lambda \right)=\sum _{k=0}^{\infty }p\left(k;\lambda \right){e}^{ik𝜃}={e}^{\lambda \left({e}^{i𝜃}-1\right)}\phantom{\rule{0.3em}{0ex}}.$

For further properties see Breiman, [1, page 170 ﬀ.], especially Deﬁnition 8.26 and Proposition 8.27, or Chung, [2, page 142 ﬀ.], especially item 6 on page 147. The characteristic function is a $2\pi$-periodic function with a series deﬁnition which converges uniformly on $ℝ$, meaning we can integrate the series term-by-term on $\left[-\pi ,\pi \right]$ to obtain

 $p\left(k;\lambda \right)=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }\stackrel{̂}{p}\left(𝜃,\lambda \right){e}^{-ik𝜃}\phantom{\rule{0.3em}{0ex}}d𝜃$ (1)

which is valid for $\lambda >0$ and $k=0,1,2,3,\dots$. This is the Fourier inversion formula for the characteristic function. See also Breiman, [1], Theorem 8.39, page 178; Chung, [2], Theorem 6.2.3; or Feller, [3], page 509.

In equation (1) set $\lambda =k$, and deﬁne ${I}_{k}$ by

${I}_{k}=\frac{{k}^{k}}{k!}{e}^{-k}=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{e}^{k\left({e}^{i𝜃}-1-i𝜃\right)}\phantom{\rule{0.3em}{0ex}}d𝜃\phantom{\rule{1em}{0ex}}k=0,1,2,3,\dots$

Compare this representation to the Gaussian integral with variance $k$ deﬁned as

${J}_{k}=\frac{1}{\sqrt{2\pi k}}=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }{e}^{-k{𝜃}^{2}∕2}\phantom{\rule{0.3em}{0ex}}d𝜃\phantom{\rule{0.3em}{0ex}}.$

The general plan is to show that ${I}_{k}\approx {J}_{k}$, so that then we can assert that

$\frac{{k}^{k}}{k!}{e}^{-k}\approx \frac{1}{\sqrt{2\pi k}}$

which can be rearranged to Stirling’s Formula. The detailed plan is to show that this approximation can be expressed as an asymptotic limit.

Break ${I}_{k}$ into pieces to make the following deﬁnitions

${I}_{k}=\frac{1}{2\pi }{\int }_{|𝜃|\le 1}{e}^{k\left({e}^{i𝜃}-1-i𝜃\right)}\phantom{\rule{0.3em}{0ex}}d𝜃+\frac{1}{2\pi }{\int }_{1<|𝜃|\le \pi }{e}^{k\left({e}^{i𝜃}-1-i𝜃\right)}\phantom{\rule{0.3em}{0ex}}d𝜃={I}_{k}^{\left(1\right)}+{I}_{k}^{\left(2\right)}$

and

${J}_{k}=\frac{1}{2\pi }{\int }_{|𝜃|\le 1}{e}^{-k{𝜃}^{2}∕2}\phantom{\rule{0.3em}{0ex}}d𝜃+\frac{1}{2\pi }{\int }_{1<|𝜃|}{e}^{-k{𝜃}^{2}∕2}\phantom{\rule{0.3em}{0ex}}d𝜃={J}_{k}^{\left(1\right)}+{J}_{k}^{\left(2\right)}$

Lemma 1. The complex exponential has the following properties and estimates:

1. For $A\in ℂ$
 $|{e}^{A}|={e}^{\Re A},$ (2)
2. For $𝜃\in ℝ$
 $|{e}^{i𝜃}-1-i𝜃|\le {𝜃}^{2}∕2,$ (3)
3. For $𝜃\in ℝ$
 $|{e}^{i𝜃}-1-i𝜃+{𝜃}^{2}∕2|\le |𝜃{|}^{3}∕3!,$ (4)
4. For $A,B\in ℂ$
 $|{e}^{A}-{e}^{B}|\le |A-B|{e}^{max\left[\Re A,\Re B\right]}.$ (5)

Proof. Left as exercises. □

Use the triangle inequality for integrals and equation (2) to bound ${I}_{k}^{\left(2\right)}$ and ${J}_{k}^{\left(2\right)}$ as

$\begin{array}{llll}\hfill {I}_{k}^{\left(2\right)}& \le \frac{1}{2\pi }{\int }_{1<|𝜃|\le \pi }|{e}^{k\left({e}^{i𝜃}-1-i𝜃\right)}|\phantom{\rule{0.3em}{0ex}}d𝜃\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \le \frac{1}{2\pi }{\int }_{1<|𝜃|\le \pi }{e}^{kcos\left(𝜃\right)-1}\phantom{\rule{0.3em}{0ex}}d𝜃\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \le {e}^{k\left(cos\left(1\right)-1\right)}\frac{1}{2\pi }{\int }_{1<|𝜃|\le \pi }\phantom{\rule{0.3em}{0ex}}d𝜃\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \le {e}^{k\left(cos\left(1\right)-1\right)}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Use a technique similar to the proof of Markov’s inequality to estimate ${J}_{k}^{\left(2\right)}$ as

${J}_{k}^{\left(2\right)}=\frac{1}{2\pi }{\int }_{|𝜃|>1}{e}^{-k{𝜃}^{2}∕2}\phantom{\rule{0.3em}{0ex}}d𝜃\le \frac{1}{2\pi }{\int }_{|𝜃|>1}|𝜃|{e}^{-k{𝜃}^{2}∕2}\phantom{\rule{0.3em}{0ex}}d𝜃=\frac{1}{2\pi }\frac{2}{k}{e}^{-k∕2}.$

Both ${I}_{k}^{\left(2\right)}$ and ${J}_{k}^{\left(2\right)}$ tend to zero at an exponential rate.

Now the eﬀort is to estimate the closeness of ${I}_{1}^{\left(1\right)}$ and ${J}_{1}^{\left(1\right)}$. Write

${I}_{k}^{\left(1\right)}-{J}_{k}^{\left(1\right)}=\frac{1}{2\pi }{\int }_{-1}^{1}\left({e}^{k\left({e}^{i𝜃}-1-i𝜃\right)}-{e}^{-k{𝜃}^{2}∕2}\right)\phantom{\rule{0.3em}{0ex}}d𝜃.$

For $|𝜃|\le 1$, use inequality (4) to derive

$\begin{array}{llll}\hfill |cos\left(𝜃\right)-1+{𝜃}^{2}∕2|& \le |cos\left(𝜃\right)+isin\left(𝜃\right)-1-i𝜃+{𝜃}^{2}∕2|\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \le |{e}^{i𝜃}-1-i𝜃+{𝜃}^{2}∕2|\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \le \frac{|{𝜃}^{3}|}{6}\le \frac{{𝜃}^{2}}{6}.\phantom{\rule{2em}{0ex}}& \hfill \text{(6)}\end{array}$

Therefore, $cos\left(𝜃\right)-1\le -{𝜃}^{2}∕3$.

Now put these together using inequalities (6) and (4)

$|{I}_{k}^{\left(1\right)}-{J}_{k}^{\left(1\right)}|\le \frac{1}{2\pi }{\int }_{-1}^{1}\left|{e}^{k\left({e}^{i𝜃}-1-i𝜃\right)}-{e}^{-k{𝜃}^{2}∕2}\right|\phantom{\rule{0.3em}{0ex}}d𝜃\le k{\int }_{-1}^{1}\frac{|𝜃{|}^{3}}{3!}{e}^{-k{𝜃}^{2}∕3}\phantom{\rule{0.3em}{0ex}}d𝜃$

Change variables with $\varphi =\sqrt{k}𝜃$ to obtain (the derivation is left as an exercise)

$\begin{array}{llll}\hfill k{\int }_{-1}^{1}\frac{|𝜃{|}^{3}}{3!}{e}^{-k{𝜃}^{2}∕3}\phantom{\rule{0.3em}{0ex}}d𝜃& =\frac{1}{6k}{\int }_{-\sqrt{k}}^{\sqrt{k}}|{\varphi }^{3}|{e}^{-{\varphi }^{2}∕3}\phantom{\rule{0.3em}{0ex}}d\varphi \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{3}{2k}-\frac{{e}^{-k∕3}}{2}-\frac{3{e}^{-k∕3}}{2k}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Then putting all these together $|{I}_{k}-{J}_{k}|\to 0$. Recalling the deﬁnitions of ${I}_{k}$ and ${J}_{k}$, this is the same as

$\left|\frac{k!{e}^{-k}}{k!}-\frac{1}{\sqrt{2\pi k}}\right|\to 0$

as $k\to \infty$. This is equivalent to Stirling’s Formula

$\underset{k\to \infty }{lim}\frac{\sqrt{2\pi k}{k}^{k}{e}^{-k}}{k!}=1.$

##### An alternate proof using the Lebesgue Dominated Convergence Theorem

Start with the deﬁnition of ${I}_{k}$

${I}_{k}=\frac{{k}^{k}}{k!}{e}^{-k}=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{e}^{k\left({e}^{i𝜃}-1-i𝜃\right)}\phantom{\rule{0.3em}{0ex}}d𝜃\phantom{\rule{0.3em}{0ex}}.$

Make the change of variables $y=𝜃\sqrt{k}$ with $\phantom{\rule{0.3em}{0ex}}dy=\phantom{\rule{0.3em}{0ex}}d𝜃\sqrt{k}$ to obtain

${I}_{k}\sqrt{k}=\frac{{k}^{k}\sqrt{k}}{k!}{e}^{-k}=\frac{1}{2\pi }{\int }_{-\pi \sqrt{k}}^{\pi \sqrt{k}}{e}^{k\left({e}^{iy∕\sqrt{k}}-1-iy\sqrt{k}\right)}\phantom{\rule{0.3em}{0ex}}dy\phantom{\rule{0.3em}{0ex}}.$

Consider the integrand ${e}^{k\left({e}^{iy∕\sqrt{k}}-1-iy\sqrt{k}\right)}$. The exponent converges pointwise

$k\left({e}^{iy∕\sqrt{k}}-1-iy∕\sqrt{k}\right)\to -{y}^{2}∕2$

as $k\to \infty$ by using equation (4), so the integrand converges pointwise to

${e}^{k\left({e}^{iy∕\sqrt{k}}-1-iy\sqrt{k}\right)}\to {e}^{-{y}^{2}∕2}\phantom{\rule{0.3em}{0ex}}.$

The integrand is bounded pointwise by

$|{e}^{k\left({e}^{iy∕\sqrt{k}}-1-iy\sqrt{k}\right)}|={e}^{k\left(cos\left(y∕\sqrt{k}\right)-1\right)}$

using equation (2). Using the half-angle identity, this can be written as

${e}^{k\left(cos\left(y∕\sqrt{k}\right)-1\right)}={e}^{-2k{sin}^{2}\left(y∕\left(2\sqrt{k}\right)\right)}\phantom{\rule{0.3em}{0ex}}.$

Finally,

${e}^{-2k{sin}^{2}\left(y∕\left(2\sqrt{k}\right)\right)}\le {e}^{-2{y}^{2}∕{\pi }^{2}}$

if and only if $-2k{sin}^{2}\left(y∕\left(2\sqrt{k}\right)\right)\le -2{y}^{2}∕{\pi }^{2}$ or

$\frac{{sin}^{2}\left(y∕\left(2\sqrt{k}\right)\right)}{{\left(y∕\sqrt{k}\right)}^{2}}\ge \frac{1}{{\pi }^{2}}$

on the domain of integration $\left[-\pi \sqrt{k},\pi \sqrt{k}\right]$. But the function $\frac{{sin}^{2}\left(y∕\left(2\sqrt{k}\right)\right)}{{\left(y∕\sqrt{k}\right)}^{2}}$ has a limit of $1∕4$ as $y\to 0$, is symmetric around $y=0$, and is decreasing on $\left(0,\pi \sqrt{k}\right)$. The minimum value of the function occurs at $\pi \sqrt{k}$ and is $1∕{\pi }^{2}$. Then using the Lebesgue Dominated Convergence Theorem, [2, page 42] or [1, page 33, Theorem 2.44]

${\int }_{-\pi }^{\pi }{e}^{k\left({e}^{i𝜃}-1-i𝜃\right)}\phantom{\rule{0.3em}{0ex}}d𝜃\to {\int }_{-\infty }^{\infty }{e}^{-{y}^{2}∕2}\phantom{\rule{0.3em}{0ex}}dy=\sqrt{2\pi }\phantom{\rule{0.3em}{0ex}}.$

Putting this together

$\frac{k!\sqrt{k}{e}^{-k}}{k!}\to \sqrt{2\pi }$

as $k\to \infty$. This is equivalent to Stirling’s Formula

$\underset{k\to \infty }{lim}\frac{\sqrt{2\pi k}{k}^{k}{e}^{-k}}{k!}=1.$

##### Discussion

These proofs establishes the Stirling’s Formula asymptotic limit fairly easily, but are not enough to show the inequality

$\sqrt{2\pi }{n}^{n+1∕2}{e}^{-n+1∕\left(12n+1\right)}

In order to establish the inequality requires bounds on the rate of approach of

${I}_{k}=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{e}^{k\left({e}^{i𝜃}-1-i𝜃\right)}\phantom{\rule{0.3em}{0ex}}d𝜃$

to

${J}_{k}=\frac{1}{\sqrt{2\pi k}}=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }{e}^{-k{𝜃}^{2}∕2}\phantom{\rule{0.3em}{0ex}}d𝜃\phantom{\rule{0.3em}{0ex}}.$

Such an estimate requires bounds on the rate of approach of

${e}^{k\left({e}^{iy∕\sqrt{k}}-1-iy\sqrt{k}\right)}\to {e}^{-{y}^{2}∕2}$

which is possible with careful estimation.

#### Sources

The heuristic proof using the Central Limit Theorem is adapted from Ross [6, pages 171-172], which in turn is based on Hu [4]. The rigorous proof is adapted from the short article by Pinsky [5].

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### Problems to Work for Understanding

1. Show that for $A\in ℂ$, $|{e}^{A}|={e}^{\Re A}$
2. Show that for $𝜃\in ℝ$, $|{e}^{i𝜃}-1-i𝜃|\le {𝜃}^{2}∕2$
3. Show that for $𝜃\in ℝ$, $|{e}^{i𝜃}-1-i𝜃+{𝜃}^{2}∕2|\le |{𝜃}^{3}∕3!|$
4. Use standard theorems from calculus (either the Fundamental Theorem of Calculus or the Mean Value Theorem) applied to the function $f\left(t\right)={e}^{tA+\left(1-t\right)B}$ to show that for $A,B\in ℂ$, $|{e}^{A}-{e}^{B}|\le |A-B|{e}^{max\left[\Re A,\Re B\right]}$
5. Show that
$\frac{1}{6k}{\int }_{-\sqrt{k}}^{\sqrt{k}}|{\varphi }^{3}|{e}^{-{\varphi }^{2}∕3}\phantom{\rule{0.3em}{0ex}}d\varphi =\frac{3}{2k}-\frac{{e}^{-k∕3}}{2}-\frac{3{e}^{-k∕3}}{2k}$

6. Derive inequalities to estimate the size of the diﬀerence
${I}_{k}-{J}_{k}=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{e}^{k\left({e}^{i𝜃}-1-i𝜃\right)}\phantom{\rule{0.3em}{0ex}}d𝜃-\frac{1}{\sqrt{2\pi k}}=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }{e}^{-k{𝜃}^{2}∕2}\phantom{\rule{0.3em}{0ex}}d𝜃\phantom{\rule{0.3em}{0ex}}.$

Use these inequalities to derive inequalities for $k!$ reﬁning Stirling’s asymptotic limit formula to an inequality.

7. In the intuitive proof, the key approximation is $\begin{array}{llll}\hfill ℙ\left[{S}_{n}=n\right]& =ℙ\left[n-1<{S}_{n}\le n\right]\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =ℙ\left[-1∕\sqrt{n}<{Z}_{n}\le 0\right]\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx {\int }_{-1∕\sqrt{n}}^{0}\frac{1}{\sqrt{2\pi }}{e}^{-{x}^{2}∕2}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx \frac{1}{\sqrt{2\pi }}\frac{1}{\sqrt{n}}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}& \hfill \end{array}$

In the Poisson probability, the interval (for example) $\left(n-2∕3,n+2∕3\right]$ could replace the interval $\left(n-1,n\right]$. But then the value of the integral is proportional to the length $4∕3$ of the interval, producing the wrong result. If (for example) $\left(n-1∕2,n+1∕2\right]$ replaces the interval $\left(n-1,n\right]$, the result is correct. How does the intuitive proof change when calculating the respective probabilities with an interval of length not equal to $1$?

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### References

[1]   Leo Breiman. Probability. Addison Wesley, 1968.

[2]   Kai Lai Chung. A Course in Probability Theory. Academic Press, 1974.

[3]   William Feller. A Introduction to Probability Theory and It Applications, Volume II, Second Edition, volume II. John Wiley and Sons, second edition, 1971.

[4]   T.-C. Hu. A statistical method of approach to Stirling’s formula. American Statistician, 42:204–205, 1988.

[5]   Mark A. Pinsky. Stirling’s formula via the Poisson distribution. American Mathematical Monthly, 114(3):256–258, March 2007.

[6]   Sheldon M. Ross. Introduction to Probability Models. Elsevier, 6th edition, 1997.

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