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probability代写| MA2506代写|概率论代写

问题 1 (problem1). A coin is tossed repeatedly until it lands Heads for the first time. Let \(X\) be the number of tosses that are required (including the toss that landed Heads), and let \(p\) be the probability of Heads. Find the CDF of \(X,\) and for \(p=1 / 2\) sketch its graph.
证明 (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 \(x<1,\).
问题 2 (problem2). There are 100 shoelaces in a box. At each stage, you pick two random ends and tie them together. Either this results in a longer shoelace (if the two ends came from different pieces), or it results in a loop (if the two ends came from the same piece). What are the expected number of steps until everything is in loops, and the expected number of loops after everything is in loops? (This is a famous interview problem; leave the latter answer as a sum.)
证明 (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 $$ \sum_{n=1}^{100} \frac{1}{2 n-1} $$
问题 3 (problem3). 7. Athletes compete one at a time at the high jump. Let \(X_{j}\) be how high the \(j\) th jumper jumped, with \(X_{1}, X_{2}, \ldots\) i.i.d. with a continuous distribution. We say that the \(j\) th jumper set a record if \(X_{j}\) is greater than all of \(X_{j-1}, \ldots, X_{1}\). (a) Is the event “the 110 th jumper sets a record” independent of the event “the 111 th jumper sets a record”? Justify your answer by finding the relevant probabilities in the definition of independence and with an intuitive explanation. (b) Find the mean number of records among the first \(n\) jumpers (as a sum). What happens to the mean as \(n \rightarrow \infty ?\)
证明 (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).
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