Help! Will give Brainliest

Why would you even ask this ._.

oee [108]

Answer:

Ask something & it will be answered

Step-by-step explanation:

7 0

3 years ago

Let X and Y be discrete random variables. Let E[X] and var[X] be the expected value and variance, respectively, of a random vari

Ulleksa [173]

Answer:

(a) $E[X+Y]=E[X]+E[Y]$

(b) $Var(X+Y)=Var(X)+Var(Y)$

Step-by-step explanation:

Let X and Y be discrete random variables and E(X) and Var(X) are the Expected Values and Variance of X respectively.

(a)We want to show that E[X + Y ] = E[X] + E[Y ].

When we have two random variables instead of one, we consider their joint distribution function.

For a function f(X,Y) of discrete variables X and Y, we can define

$E[f(X,Y)]=\sum_{x,y}f(x,y)\cdot P(X=x, Y=y).$

Since f(X,Y)=X+Y

$E[X+Y]=\sum_{x,y}(x+y)P(X=x,Y=y)\\=\sum_{x,y}xP(X=x,Y=y)+\sum_{x,y}yP(X=x,Y=y).$

Let us look at the first of these sums.

$\sum_{x,y}xP(X=x,Y=y)\\=\sum_{x}x\sum_{y}P(X=x,Y=y)\\\text{Taking Marginal distribution of x}\\=\sum_{x}xP(X=x)=E[X].$

Similarly,

$\sum_{x,y}yP(X=x,Y=y)\\=\sum_{y}y\sum_{x}P(X=x,Y=y)\\\text{Taking Marginal distribution of y}\\=\sum_{y}yP(Y=y)=E[Y].$

Combining these two gives the formula:

$\sum_{x,y}xP(X=x,Y=y)+\sum_{x,y}yP(X=x,Y=y) =E(X)+E(Y)$

Therefore:

$E[X+Y]=E[X]+E[Y] \text{ as required.}$

(b)We want to show that if X and Y are independent random variables, then:

$Var(X+Y)=Var(X)+Var(Y)$

By definition of Variance, we have that:

$Var(X+Y)=E(X+Y-E[X+Y]^2)$

$=E[(X-\mu_X +Y- \mu_Y)^2]\\=E[(X-\mu_X)^2 +(Y- \mu_Y)^2+2(X-\mu_X)(Y- \mu_Y)]\\$Since we have shown that expectation is linear$\\=E(X-\mu_X)^2 +E(Y- \mu_Y)^2+2E(X-\mu_X)(Y- \mu_Y)]\\=E[(X-E(X)]^2 +E[Y- E(Y)]^2+2Cov (X,Y)$

Since X and Y are independent, Cov(X,Y)=0

$=Var(X)+Var(Y)$

Therefore as required:

$Var(X+Y)=Var(X)+Var(Y)$

7 0

3 years ago

In square ABCD, m∠AED=(7x−8)°. What is the value of x?

kirill115 [55]

Answer:

12

Step-by-step explanation:

The angle equals 90 degrees so (8*12-6) would equal 90

hopefully i helped

4 0

3 years ago

The area of a regular hexagon circumscribed about a circle with radius 2 is _______________.

Yuliya22 [10]

Answer:

relate the values and find the correct answer

7 0

3 years ago

Read 2 more answers

ERROR ANALYSIS In Exercises 39 and 40, describe and

expeople1 [14]

Answer:

39. r = 13.4

40. m = 12

Step-by-step explanation:

39. Given the equation, $-0.8 + r = 12.6$ , to solve for r, the following are the correct steps to take to arrive at the solution:

$-0.8 + r = 12.6$ (given)

Add 0.8 to both sides of the equation (addition property of equality)

$-0.8 + r + 0.8 = 12.6 + 0.8$ (this is where the error occurred.)

$r = 13.4$

40. The correct steps to take in solving the equation, $-\frac{m}{3} = -4$ is as follows:

$-\frac{m}{3} = -4$ (given)

Multiply both sides by 3 (multiplication property of equality)

$3*-\frac{m}{3} = 3*(-4)$

$-m = -12$ (this is where the error occurred. This is what we should have at this line/step)

$m = 12$ (dividing both sides by -1)

6 0

4 years ago