Expectations and probability

Expect linearity

\(E(x+y)=E(x)+E(y)\) 任意x,y
\[E(x+y)=\sum_i \sum_j P(x=i,y=j)(i+j)\]
\[\sum_i \sum_j P(x=i,y=j) i\]

\[=\sum_i i P(x=i)\]

Similarly j

To give
\ [\ sum_iiP (x = i ) + \ sum_jjP (x = j) \]

\[=\sum_i \sum_j P(x=i,y=j) i+\sum_i \sum_j P(x=i,y=j) j\]

\[\sum_i \sum_j P(x=i,y=j)(i+j)\]

\ [= E (x + y) \]

Does not exceed the desired maximum Y

\[E(y)=\sum_i i P(y=i)\]

\[=\sum_i i (P(y<=i) - P(y<=i-1))\]

\[=\sum_i i((\frac{i}{s})^n-(\frac{i-1}{s})^n)\]

Probability \ (P \) , \ (\ FRAC. 1} {P} {\) occurring after time

Let X number

That proof \ (E (x) = \ frac {1} {p} \)

解一、\[P(x=i)=(1-p)^{i-1}*p\]
解二、\[P(x>=i)-P(x>=(i+1))=(1-p)^{i-1}*p\]

Only a few times before the tubes have failed

\[P(x>=i)= (1-p)^{i-1}\]

\[\sum _{i=1}^{inf} ((1-p)^{i-1} -(1-p)^{i})*i\]

For each \ ((. 1-P) ^ I \) , only two coefficients a i, i + 1 a coefficient
that is equal to the original formula

\[\sum_{i=0}^{inf} (1-p)^{-i}\]

\[\frac{1}{1-(p-1)}\]

\[=\frac{1}{p}\]

Ball

Back and without replacement

\ [E (s) = \ sum _i E (xi) = E (\ sum _i yi * i) = \ sum _iE (yi * i) \]

That

\[E(s)=T*\sum_i E(i) = \frac{m*(n+1)}{2}\]

\ [T = \ sum E (
yi) = m \] Perceptual understood: there is no difference between each ball

Random Walk

First find the desired questions do equivalence point, set a number of issues, to find the relationship between the different points E (next push only) solving the equation
or set one of the all cases push the end

1. Walk on the chain, from one end to the other end of a desired number of steps
   \ [E (Y) = \ {i = SUM _ ^ {}. 1. 1-n-Xi} \] Xi is a random walk first come i i +1 step number
   \ [E (Y) = \ {I SUM _ =. 1} ^ {}. 1-n-E (Xi) \]
   \ [E (X2) =. 1 / + 2. 1/2 * (. 1 + E (the X1) + E (an X2)) \]
   \ [E (XI) =. 1 / + 2. 1/2 * (. 1 + E (. 1-Xi) + E (Xi)) = E (Xi-. 1) +2 \]
   \ [E (Y) = \ {I SUM _ =. 1} ^ {}. 1-n-E (Xi) =. 5. 1 + + +. 7. 3. 9 + ... + = (. 1-n-) ^ 2 \ ]
2. The complete graph walk, a few steps to the desired point to another point
   \ (1 / n-1 \ ) the probability of success (every time to come every other point)
   with a coin toss about the same, that is, each \ (1 / n-1 \) thrown into the front, is the expected number of steps n-1
3. 2n point walk on complete bipartite graph, a point to another desired number of steps
   A: the number of steps to the ipsilateral point
   B: opposite side point
   \ (B = 1 / n + 1 + n-1 / n * (B + 2) \)
   \ (A = (. 1 + B) \)
4. FIG walk in chrysanthemum n, the number of steps is desirable to the root of x
   \ (. 1 = E / (. 1-n) + (n-2) / (. 1-n) * (2 + E) \)
   \ (E / ( . 1-n-) = (. 1-2N) / (. 1-n-) \)
   \ (E = (2N-. 1) \)
5. tree walk point n, x-> y a desired number of steps required
   to root y is
   f (x) from x to x father first
   D [x] x degree
   \ (f (x) = 1 / d [x] + 1 / d [x] *
   \ sum_ {i-1} ^ {d [x], i! = fa [x]} (1 + f (son [x] + f (x)) \) then blind JB engage in DP
6. 200 points configured undirected graph, so that the desired S to T \ (\ ge1000000 \)
   E1 If the O (n ^ 2), the chain 100 is engaged

Classic problem

• Each time a random integer ⼀ [1, n], and asked to get accustomed desired number of times can scrape all ⻬
  \ [\ sum_i n / i \
  ] each have disposed out \ (n-1 \) a, not drawn to smoked the probability of success \ ((n-I-+. 1) / n-\) , a desired number of times \ (n / (n-i + 1) \)
• ⻓ ⼀ a random arrangement of the n-p, find p [1 ... i] in the p [i] is the probability is largest number of
  \ [1 / i \]
• ⾯ on the question asked ⽅ full underexposure level of a desired number of i
  \ [\ sum_ {i! = J} 1 / ij + \ sum_i 1 / i \]
• ⻓ ⼀ a random arrangement of the n-p, i in the determination of probability of the screen for j
  \ [\ frac {1} { 2} \]
• Random ⼀ a ⻓ degree n of arrangement p, find it contains w [1 ... m] probability of a sequence as slave sequence / continuous Submenu
  \ [\ frac {1} { m!} * (N-m + 1) \]
  \ [\ FRAC. 1 {m} {!} * n_m C ^ \] (presumably?

• N-⽯ head stack, the stack number of the i A [i], each a randomly selected ⼀ ⽯ head and put the whole stack are thrown ⼀, first find the desired stack ⽯ head of times to get accustomed thrown
  \ [1 + \ sum_ {i = 2} ^ nP (A [i] <A [1]) \]

  \(P(A[i]<A[1])=a[i]/(a[i]+a[1])\)
• A random ⼀ ⻓ string length n is 01, each position is a probability p, defined and X is ⽅ of each successive level of 1 degree and ⻓, demand E [X]

• To ⼀ sequences, each randomly delete ⼀ elements, ask the probability of i and j are adjacent in the process,

Namely \ (i ~ j \) number of permutations intermediate, i, j in the last program number
\ [(. 1-JI)! * 2 / (+ JI. 1)! \]
\ [= 2 / (JI ) * (j-i + 1 ) \]

• Given ⼀ tree, his random side Insert sequentially after sequential ⼀, seeking u, v when the desired communication
• 1 ... n n to this number, each time to choose an random number and also deleted his number all about, seeking expect to get accustomed finished second deletion