Our inductive hypothesis is ##\prod\limits_{n=1}^{k}n(2k+2-2n)=2^k(k!)^2##, for some ##k\in\{1,2,...\}##.
We want to show that ##\prod\limits_{n=1}^{k+1}n(2(k+1)+2-2n)=2^{k+1}((k+1)!)^2##.
I think the first equality should be ##E(X(s)X(s+t))=(\mu^{(s)})_1 p_{11}(t)##.
Note that the state space of the Markov chain is ##\{0,1\}##.
By the definition of expectation, we have ##E(X(s)X(s+t))=\sum\limits_{i=0}^{1}\sum\limits_{j=0}^{1}ijP(X(s)=i, X(s+t)=j)=P(X(s)=1, X(s+t)=1)##.
We...
To show that $$\max\limits_{n\in[1,2]}P(X(n)>x)\leq P(\max\limits_{n\in[1,2]}X(n)>x)$$, I would suggest first showing that $$P(X(n)>x)\leq P(\max\limits_{m\in[1,2]}X(m)>x) \text{ , for all } n\in[1,2]$$. (To avoid confusion, I used the variable m, instead of n, on the right hand side of this...
Denote the 3 vertices of the isosceles triangle by A, B, and C, with sides AB and BC having the same length.
Drop a perpendicular from B to AC, and denote the point of intersection by D. Note that we have BD = x.
We shall label the centre of the incircle as O, and shall drop another...
You may want to approach the problem by proving these 2 inequalities :
1) inf(A) <= -sup(-A) , or equivalently, sup(-A) <= -inf(A)
2) inf(A) >= -sup(-A)
Combining these 2 inequalities will give you the desired result.
I believe changing the order of the terms is going to have an effect on the sum at least. Suppose (x_{1},x_{2},...,x_{n}) = (1, 2, ..., n) and (y_{1},y_{2},...,y_{n}) = (n, ..., 1). In general, \sum_{j\displaystyle{=}1}^n jx_{j} is not equal to \sum_{j\displaystyle{=}1}^n jy_{j} .
Here's...
Take for example the permutation (x_{1},x_{2},...,x_{n}) = (1, 2, ..., n) . We have \sum_{j=1}^n j x_j = 1^2 + 2^2 + ... + n^2 .
Thus, the question is asking us to show that the probability of a permutation (x_{1},x_{2},...,x_{n}) having the property \sum_{j\displaystyle{=}1}^n jx_{j}...
A permutation is simply an arrangement of the numbers 1, 2, 3, ..., n. So, examples of permutations are (1, 2, 3, ..., n-1, n) ,
(2, 3, 4, ..., n, 1) etc. There are n! such permutations in all, as mentioned in the question.
For starters, there may be a slight error in the question. I think the result to be proven should be P\displaystyle{\{} \sum_{j\displaystyle{=}1}^n jX_{j} \leq a \displaystyle{\}} = P\displaystyle{\{} \sum_{j\displaystyle{=}1}^n jX_{j} \geq \frac{n(n+1)^2}{2}-a \displaystyle{\}} .
Since...
At each replication, a ball is drawn from the urn. The colour of the ball is noted, then that ball, together with a new ball of the same colour, are placed into the urn. Note then that after each replication, the number of white balls in the urn either stays the same or increases by one.
Now...
1) After the nth replication, there are Z_n(n+2) white balls and (1-Z_n)(n+2) black balls in the urn.
2) In the (n+1)th replication, there will either be a white ball or a black ball added into the urn. For each of these cases, what will the new proportion of white balls in the urn be and...
Let Y_{n} be the gambler's winnings after n games. Clearly, Y_{n} is a martingale. We introduce a new stochastic process Z_{n}, where Z_{n}={Y_{n}}^2-n. It can be shown that Z_{n} is a martingale with respect to Y_{n}. (Can you try to show this?)
Let N be the random variable for the step...