Exercises

Jun 15'24

Prove that

[[math]] u^{(2)}_{2n} = \frac 1{4^{2n}} \sum_{k = 0}^n \frac {(2n)!}{k!k!(n-k)!(n-k)!}\ , [[/math]]

and

[[math]] u^{(3)}_{2n} = \frac 1{6^{2n}} \sum_{j,k} \frac {(2n)!}{j!j!k!k!(n-j-k)!(n-j-k)!}\ , [[/math]]

where the last sum extends over all non-negative [math]j[/math] and [math]k[/math] with [math]j+k \le n[/math]. Also show that this last expression may be rewritten as

[[math]] \frac 1{2^{2n}}{{2n}\choose n} \sum_{j,k} \biggl(\frac 1{3^n}\frac{n!}{j!k!(n-j-k)!}\biggr)^2\ . [[/math]]

ABy Admin

Jun 15'24

Prove that if [math]n \ge 0[/math], then

[[math]] \sum_{k = 0}^n {n \choose k}^2 = {{2n} \choose n}\ . [[/math]]

Hint: Write the sum as

[[math]] \sum_{k = 0}^n {n \choose k}{n \choose {n-k}} [[/math]]

and explain why this is a coefficient in the product

[[math]] (1 + x)^n (1 + x)^n\ . [[/math]]

Use this, together with Exercise, to show that

[[math]] u^{(2)}_{2n} = \frac 1{4^{2n}}{{2n}\choose n}\sum_{k = 0}^n {n \choose k}^2 = \frac 1 {4^{2n}} {{2n}\choose n}^2\ . [[/math]]

ABy Admin

Jun 15'24

Using Stirling's Formula, prove that

[[math]] {{2n}\choose n} \sim \frac {2^{2n}}{\sqrt {\pi n}}\ . [[/math]]

ABy Admin

Jun 15'24

Prove that

[[math]] \sum_{j,k} \biggl(\frac 1{3^n}\frac{n!}{j!k!(n-j-k)!}\biggr) = 1\ , [[/math]]

where the sum extends over all non-negative [math]j[/math] and [math]k[/math] such that [math]j + k \le n[/math]. Hint: Count how many ways one can place [math]n[/math] labelled balls in 3 labelled urns.

ABy Admin

Jun 15'24

Using the result proved for the random walk in [math]{\mathbf R}^3[/math] in Example, explain why the probability of an eventual return in [math]{\mathbf R}^n[/math] is strictly less than one, for all [math]n \ge 3[/math]. Hint: Consider a random walk in [math]{\mathbf R}^n[/math] and disregard all but the first three coordinates of the particle's position.