probability代写| MA2506代写|概率论代写

Proof (problem1). By the story of the Geometric, we have \(X-1 \sim\) Geometric \((p)\). Using this or directly, the PMF is \(P(X=k)=p(1-p)^{k-1}\) for \(k \in\{1,2,3, \ldots\}\) (and 0 otherwise). The CDF can be obtained by adding up the PMF (from \(k=1\) to \(k=\lfloor x\rfloor,\) where \(\lfloor x\rfloor\) is the greatest integer less than or equal to \(x) .\) We can also see directly that $$ P(X \leq x)=1-P(X>x)=1-(1-p)^{\lfloor x\rfloor} $$ for \(x \geq 1,\) since \(X>x\) says that the first \(\lfloor x\rfloor\) flips land tails. The CDF is 0 for \(x<1 .\) For a fair coin, the \(\mathrm{CDF}\) is \(F(x)=1-\frac{1}{2\lfloor x\rfloor}\) for \(x \geq 1,\) and \(F(x)=0\) for \(x<1,\).

Proof (problem2). Initially there are 200 free ends. The number of free ends decreases by 2 each time since either two separate pieces are tied together, or a new loop is formed. So exactly 100 steps are always needed. Let \(I_{j}\) be the indicator \(r . v\). for whether a new loop is formed at the \(j\) th step. At the time when there are \(n\) unlooped pieces (so \(2 n\) ends), the probability of forming a new loop is \(\frac{n}{\left(\begin{array}{c}2 n \\ 2\end{array}\right)}=\frac{1}{2 n-1}\) since any 2 ends are equally likely to be chosen, and there are \(n\) ways to pick both ends of 1 of the \(n\) pieces. By linearity, the expected number of loops is $$ \sum_{n=1}^{100} \frac{1}{2 n-1} $$

Proof (problem3). Let \(I_{j}\) be the indicator r.v. for the \(j\) th jumper setting a record. By symmetry, \(P\left(I_{j}=1\right)=1 / j\) (as all of the first \(j\) jumps are equally likely to be the highest of those jumps). Also, $$ P\left(I_{110}=1, I_{111}=1\right)=\frac{109 !}{111 !}=\frac{1}{110 \cdot 111} $$ since having the 110 th and 111 th jumps both being records is the same thing as having the highest of the first 111 jumps being in position \(111,\) the second highest being in position \(110,\) and the remaining 109 being in any order. So $$ P\left(I_{110}=1, I_{111}=1\right)=P\left(I_{110}=1\right) P\left(I_{111}=1\right) $$ which shows that the 110 th jumper setting a record is independent of the 111 th jumper setting a record. Intuitively, this makes sense since learning that the 111 th jumper sets a record gives us no information about the “internal” matter of how the first 110 jumps are arranged amongst themselves. By linearity, the expected number of records among the first \(n\) jumpers is \(\sum_{j=1}^{n} \frac{1}{j}\), which goes to \(\infty\) as \(n \rightarrow \infty\) (as this is the harmonic series).