2 Probability Theory Topics Thank you! Seeking Answers!!

Upstairs, you underestimate Crouching Tiger, Hidden Dragon, and this question is not professional at all.

1.∫(0~1)∫(0~1) c/(1+x)(1+y) dxdy=1c =1cln2*ln2=1

c=1/(ln2)²

fx(x)=c∫(0~1)1/(1+x)(1+y) dy[c/(1+x)]ln2

1/[(1+x)ln2]

Similarly. fy(y)=1 [ln2(1+y)]fx(x)*fy(y)=1 [(ln2) (x+1)(1+y)]=p(x,y) so independent.

p(x=1)= p(x=2)=

p(y=0)= p(y=3)=

p(x=1,y=0)=

p(x=1,y=3)=

p(x=2,y=3)=

p(x=2,y=0)=

After observation, p(x=a)p(y=b)=p((x,y)=(a,b)), xy is not independent.

e(x)=e(y)=3*

e(xy)=3*

cov(xy)=e(xy)-e(x)e(y)=e(x²)=

e(y²)=9*

var(x)=e(x²)-e(x)²=

var(y)=e(y²)-e(y)²=

pxy=cov(x,y) root(varxvary)=root(root(54)=1 root6=root6 6

Brother, how can you ask such a professional question? Find a forum and ask.

Question 1: Meaning: Let the probability of event A occurring be and the probability of event b occurring is; If we know which birth of A is born, then the probability of B occurring is to find the probability that A will not occur. The horizontal bar means complement.

Analysis: By definition, p(b|a) = p(ab) p(a), so p(ab) = p(b|a)*p(a)=;p(ab) is the probability that a and b occur at the same time, and the probability of event a occurring is equal to the probability that a and b occur at the same time plus the probability that a and b do not occur, so the probability value of the problem is equal to p(a)-p(ab)=

Question 2: From the perspective of the number of rainy days, there are three situations, that is, it does not rain for two days, it rains on one day, and it rains on both extinct days. The question requires the probability of not raining for at least one day, that is, the sum of the probability of not raining for one day and not raining for two days, that is, the probability of rain for 1-2 days, which is equal to.

(1) e(y)=e(x1+x2-xn 2)=[e(x1)+e(x2)-e(xn)] 2=( 2, not an unbiased estimator.

2)e(x–)=e(x1+x2+..xn/n)=μd(x–)=d(x1+x2+..xn n) = n, which is an unbiased estimator.

3) e(z)=e(2x1+x2-x3 2) = d(z)=d(x1)+1 4d(x2)+1 4d(x3)=3 2 is an unbiased estimator.

The estimator x- is the most effective unbiased estimator.

The process is as follows:

Both problems are both moment estimators.

Question 1: Observe the function given by the question, there is only one unknown parameter u, and the relationship between the overall moment and the parameter should be established, so the expectation of the function should be found. The expected value e(x)=1+u can be solved by using the expectation formula for continuous random variables.

Then, the sample moment, i.e., the sample mean, is used instead of the expected value of the population moment, e(x), which is so that e(x) = the sample mean.

That is, 1+U = sample mean.

The third step is to solve u, i.e., u = sample mean -1. This equation is the estimator of the moment that is sought.

The first and second questions are based on the observed values of the given samples, calculate the mean value of the samples, and then substitute them into the formula of the first question to solve u=307 3.

In the second question, there is only one moment estimator given in the question, and the relationship between the moment estimator and the population moment should be established, so the expectation of the discrete random variable should be obtained. Derive expectation = -4 unknown parameter +3.

The sample moment is then replaced by the overall moment. i.e. e(x) = sample mean. The equation for the moment estimator is obtained, i.e., -4 parameter + 3 = sample mean.

Then, according to the above formula, the moment estimator is solved, and the moment estimator = -1 4 sample mean + 3 4.

That is, the desired result.

The process is as follows:

This question is divided into two questions, the first is to find the moment estimator, and the second is to find the moment estimate.

Estimating quantities is broadly divided into three steps:

First, find the relationship between the population moment and the sample parameters.

The sample moment is used instead of the population moment to obtain the equation about the moment estimator;

A system of equations for solving moment estimators.

Let's take requestioning as an example: first of all, there is only one unknown parameter in the question, so we can use expectations to make the overall moment. According to the expectation formula of continuous random variables, the solvable expectation e(x)=1+u.

The sample moment is then used instead of the population moment, i.e., the sample mean is used instead of the population expectation, so that the equation about the moment estimator is obtained.

Then the moment estimator is solved, and the final result is u = sample mean -1.

The answer to the second question is based on the first question. That is, the mean value of the sample to be given to the sample observation is obtained and substituted into the equation of the moment estimator obtained in the first question. The estimated moment of the solution is equal to 310 3.

To collect the first one, it takes 1 time.

Collect the second one, the probability is 5 6, it takes 6 5 times.

Collect the third one, the probability is 4 6, it takes 6 4 times.

Collect the fourth card, the probability is 3 6, it takes 6 3 times.

Collect the fifth card, the probability is 2 6, it takes 6 2 times.

Collect the sixth card, the probability is 1 6, it takes 6 1 times.

So theoretically, a total of 1+6 5+6 4+6 3+6 2+6 1=times need a gold coin.

If you need at least x cards to complete the estimated income, then.

1+x/（x-1）+x/（x-2）+·x/3+x/2+x/1≥300/10

Get x 11 (you can take 10 and try to bring it in, and in the end, so 11 will definitely be satisfied) so design at least 11 cards.

Finished! ~

Set k purchases to collect all the cards, obviously k 6

Think: The last time you draw a card, the first k-1 time to collect the other 5 cards, so p=6 * a(k-1,5) *5 (k-6) 6 k

The mathematical expectation is that the gold coin is e = p*10k (k 6) and this e should have a convergence limit, but I won't ask for :-(

Let ai, i= 1,2,..6, is the state of having a total of i cards in the hand. Each time you buy a card, the number of cards you have may change. Let m be a card once, then there is.

MA1 = 1 6A1 + 5 6A2, i.e.: If there is only one card in the hand before taking it. After taking, there is a probability of 1 6 to draw an existing card, so you still only have 1 card, and 5 6 has a probability of getting a new card, so you have two cards.

In the same way, there is: ma2 = 2 6a2 + 4 6a3, ma3 = 3 6a3 + 3 6a4, ma4 = 4 6a4 + 2 6a5, ma5 = 5 6a5 + 2 6a6, ma6 = a6, so m can be regarded as a linear transformation in the resulting linear space.

After taking it once, it reaches a1, i.e. x=(1,0,0,0, and after taking n+1 time, it reaches m n(x)Let yn be the value of the 6th coordinate of m n(x). then (yn-y(n-1)) is the probability of getting the 6th card exactly n+1 time.

The expected value of the required number of sheets is: (yn-y(n-1))*n+1) for n=5,6,..of sums.

The amount of calculation is relatively large, I calculated:

yn= -(1/6)^n + 5(2/6)^n - 10(3/6)^n + 10(4/6)^n - 5/6)^n + 1

The next calculation involves summing the series in the form of na n, do the math yourself.

If you don't know how to do it, you can keep asking.

The second question is to change the number of types of cards so that the expected value of the first question exceeds 300 10 = 30You can change the 6 of the first question to the general n to deal with it.

Let the train body be [0,100], and let the first shell fall at point x. Let the second shell land at point y. Let the third shell fall at point z.

0<=y<=x<=z<=100

To meet the requirements, you must have: 20<= x <= 80, 10<= y <= x-10, x+10<=z<=90

Given x, the probability that the second shell is effective is = the effective value of y and the possible value of y = (x-20) x

The probability that the third shell is effective is = the effective value of z and the possible value of z = (80-x) (100-x).

The probability is the integral of x as follows:

1 100 points (x-from 20 to 80) (x-20) x) *80-x) (100-x)) dx

1 100 points (x from 20 to 80) (1 - 16 x - 16 (100-x)) dx

60-16ln4-16ln4)/100= (15-16ln2)/25

Thanks to the Divine Insult for the reminder, I have corrected the calculation error of the last step of the points.

Three-dimensional geometric generalizations.

The four sections of the last train are marked as x, y, z, 100-x-y-z, so the sample space is x, y, z, 100-x-y-z, all of which are greater than zero, and their corresponding volumes are 100 3 6

The corresponding events are that x, y, z, and 100-x-y-z are all greater than 10, and their corresponding volumes are 80 3 6

So the probability is (