Two Probability Theory Fill in the Blank Questions Thank you, masters

I'll give you a detailed description of the process and related formula knowledge points.

1. The test is the variance of the Poisson distribution, some people say that it should be pi, because different textbooks use different letters, all of which represent the Poisson distribution.

The basic properties of the Poisson distribution p( ): e(x)=d(x)= , (is a parameter, this can be used directly if you remember this).

So for this question it is d(x)=2

In addition, the basic formula for expected variance is (where a and b are constants, and the two variables are shown in question 2):

e(ax+b)=ae(x)+b,d(ax+b)=a d(x) (this is often used, be sure to remember clearly).

So there is d(2x-3)=4d(x)=4*2=8

2. For two variables.

e(x±y)=e(x)±e(y)

d(x±y)=d(x)+d(y)±2cov(x,y)

cov(x,y) = e(xy)-e(x)e(y) This is called covariance.

Since those two variables are known to be independent of each other, there is e(xy) = e(x)e(y), so the covariance cov(x,y) = 0, so d(x y) = d(x) + d(y).

d(3x-2y)=3 d(x)+2 d(y)=44 for the original question

Question 1: p(2) find d(2 -3).

Seek d( ) first

Since: p(2).

Then: p( =k)=2 k k!*exp(-2)d(ε)=e(ε^2)-e(ε)

e(ε)=∑k*2^k/k!*exp(-2)=2∑2^(k-1)/(k-1)!*exp(-2)=2

e(ε^2)=∑k^2*2^k/k!*exp(-2)=∑k*(k-1)*2^k/k!*exp(-2)+∑k*2^k/k!

exp(-2)=4∑2^(k-2)/(k-2)!*exp(-2)+2∑2^(k-1)/(k-1)!*exp(-2)=6

d(ε)=6-2^2=2

d(2ε-3)=4d(ε)=8

The second question: and independent of each other, then cov( ,=e( e( )e( )e( )=0d(3 -2 )=d(3 )+d(2)-2cov(3,2)9d( )4d( )12cov( ,=9*4+4*2=36+8=44

The first question p is pi.

2。So d(x)=2, d(2x-3)=4d(x)+d(3)=4*2+0=8

Question 2: d(3x-2y)=9d(x)+4d(y)=44 should be...

<> fruit is not necessarily right, and this is the case with the square Piperson burning mu method.

Hello friends, Lu Mu! Finish the detailed dress to accompany the clear process of the lack of forest, I hope to help you solve the problem.

If you're not mistaken, it should be that Xun Bo Yin is a mu hall like Yinsun.

1(1): The total possible case is 5 4, one box has a ball 5, two boxes have a ball 10*(2 4- 2), the probability is 145 625 = 29 125

1 (2): The total possible situation is 5 4, one empty box is 5*4*3*2*1, and two empty boxes 10*6*3*2*1 probability is 96 125

2: There is always a possible situation of 10 3, and the probability of 5 3+5 cannot be arranged in three odd numbers is 1-13 100=87 100

According to the nature of the probabilistic law, there is p(x=k)=1. c attack ( k ) ( k!)=1，k=1,2，…, again, by the Taylor series of e x, there is e x= (x k) (k!).)【k=0,1,2,…，=1+∑(x^k)/(k!)【k=1,2……, let x= . ∴k)/(k!)=e^λ-1。∴c=1/(e^λ-1)。

FYI.

Take the logarithm of l and then find the derivative of p, and find the extreme condition of log(l).

Simple calculations, conversion of variance expressions to in-question expressions. If you want to save time, you can also make n=2, and you can directly determine the answer.

Web Links.

The moment estimation of theta is carried out by substituting the origin moment e(x) with xbar, and thetabar=2*xbar-5 is obtained

The answer has to do with the exponential distribution notation. The Economic Math team will help you find out. Please pay the ** price. Thank you!