Quelle est la différence attendue entre le maximum et le minimum de 4 variables aléatoires discrètes avec une distribution uniforme?


Réponse 1:

Définitions:

nfriends([math]i{1,...,n}[/math])eachtossa[math]k[/math]sideddie(whosevaluestakeauniformdistributionover[math]{1,...,k}[/math]).Defineeachofthe[math]n[/math]tossesresultstobearandomvariable[math]Xi{1,..,k}[/math].Definetherandomvariables[math]Xmax=max(X1,...,Xn)[/math]and[math]Xmin=min(X1,...,Xn)[/math].Wewanttofind[math]E[XmaxXmin][/math].Forourspecificquestion,[math]n=4[/math]and[math]k=6[/math].n friends ([math]i\in\{1,...,n\}[/math]) each toss a [math]k[/math]-sided die (whose values take a uniform distribution over [math]\{1,...,k\}[/math]). Define each of the [math]n[/math] tosses' results to be a random variable [math]X_i\in\{1,..,k\}[/math]. Define the random variables [math]X_{max} = \max{(X_1, ..., X_n)}[/math] and [math]X_{min} = \min{(X_1, ..., X_n)}[/math]. We want to find [math]E[X_{max}-X_{min}][/math]. For our specific question, [math]n=4[/math] and [math]k=6[/math].

Répondre:

E[XmaxXmin]=k12kn[ζ(n)ζ(n,k)],where[math]ζ(i)[/math]isRiemannZetaFunctionand[math]ζ(i,j)[/math]isHurwitzZetaFunction.E[X_{max}-X_{min}] = \boxed{k-1-\frac{2}{k^n}[\zeta(-n)-\zeta(-n,k)]}, where [math]\zeta(i)[/math] is Riemann Zeta Function and [math]\zeta(i,j)[/math] is Hurwitz Zeta Function.

Forourspecificquestion,61264[ζ(4)ζ(4,6)]=22616483.49.For our specific question, 6 - 1 - \frac{2}{6^4}[\zeta(-4) - \zeta(-4,6)] = \boxed{\frac{2261}{648}} \approx 3.49.

Raisonnement:

First,wegetP(Xmaxx)=P(X1x,...,Xnx).Assumingthateachdietossisindependentoftheothertosses,[math]P(X1x,...,Xnx)=i=1nP(Xix)=(xk)n[/math].Then,notingthat[math]P(Xmax=x)=P(Xmaxx)P(Xmaxx1)[/math],wegettheprobabilitymassfunction[math]P(Xmax=x)=(xk)n(x1k)n[/math].First, we get P(X_{max} \leq x) = P(X_1 \leq x, ..., X_n \leq x). Assuming that each die toss is independent of the other tosses, [math]P(X_1 \leq x, ..., X_n \leq x) = \prod\limits_{i=1}^{n}{P(X_i \leq x)} = (\frac{x}{k})^n[/math]. Then, noting that [math]P(X_{max} = x) = P(X_{max} \leq x) - P(X_{max} \leq x-1)[/math], we get the probability mass function [math]P(X_{max} = x) = (\frac{x}{k})^n - (\frac{x - 1}{k})^n[/math].

Similarly,wegetP(Xminx)=P(X1x,...,Xnx).Againassumingtheindependenceofdietosses,[math]P(X1x,...,Xnx)=i=1nP(Xix)=(kx+1k)n[/math].Then,notingthat[math]P(Xmin=x)=P(Xminx)P(Xminx+1)[/math],wegettheprobabilitymassfunction[math]P(Xmin=x)=(kx+1k)n(kxk)n[/math].Similarly, we get P(X_{min} \geq x) = P(X_1 \geq x, ..., X_n \geq x). Again assuming the independence of die tosses, [math]P(X_1 \geq x, ..., X_n \geq x) = \prod\limits_{i=1}^{n}{P(X_i \geq x)} = (\frac{k-x+1}{k})^n[/math]. Then, noting that [math]P(X_{min} = x) = P(X_{min} \geq x) - P(X_{min} \geq x+1)[/math], we get the probability mass function [math]P(X_{min} = x) = (\frac{k - x + 1}{k})^n - (\frac{k - x}{k})^n[/math].

Nowthatwehavebothrandomvariablesprobabilitymassfunctions,wecansimplyuselinearityofexpectedvaluetocomputeE[Xmax]E[Xmin]insteadof[math]E[XmaxXmin][/math].Bythedefinitionofexpectedvalueforadiscreterandomvariable,[math]E[Xmax]=x=1kxP(Xmax=x)=x=1kx[(xk)n(x1k)n][/math].Similarly,[math]E[Xmin]=x=1kxP(Xmin=x)=x=1kx[(kx+1k)n(kxk)n][/math].Subtracting[math]E[Xmin][/math]from[math]E[Xmax][/math]andsimplifying,weget[math]kn+1kn2x=1k1xnkn[/math],whichsimplifiestotheexpressioninAnswer(since[math]x=1k1xn=ζ(n)ζ(n,k)[/math]).Now that we have both random variables' probability mass functions, we can simply use linearity of expected value to compute E[X_{max}]-E[X_{min}] instead of [math]E[X_{max}-X_{min}][/math]. By the definition of expected value for a discrete random variable, [math]E[X_{max}] = \sum\limits_{x=1}^{k}{xP(X_{max} = x)} = \sum\limits_{x=1}^{k}{x[(\frac{x}{k})^n - (\frac{x - 1}{k})^n]}[/math]. Similarly, [math]E[X_{min}] = \sum\limits_{x=1}^{k}{xP(X_{min} = x)} = \sum\limits_{x=1}^{k}{x[(\frac{k - x + 1}{k})^n-(\frac{k - x}{k})^n]}[/math]. Subtracting [math]E[X_{min}][/math] from [math]E[X_{max}][/math]and simplifying, we get [math]\frac{k^{n+1}-k^n-2\sum\limits_{x=1}^{k-1}{x^n}}{k^n}[/math], which simplifies to the expression in Answer (since [math]\sum\limits_{x=1}^{k-1}{x^n} = \zeta(-n)-\zeta(-n,k)[/math]).