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musickatia [10]
3 years ago
15

Please solve it as fast as possible​

Mathematics
1 answer:
Alja [10]3 years ago
3 0
Here’s the answer, i hope this helped!

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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
Besides -10.75+-10.75 what plus what = -21.5?
Butoxors [25]

Answer:

-32.25 + 10.75 = -21.5

Step-by-step explanation:

-10.75 + -10.75  = -21.5

-32.25 + 10.75 = -21.5

5 0
3 years ago
Read 2 more answers
One of the angles of a regular dodecagon is 150°. What is the sum of the measures of all the interior angles of this dodecagon?
Kamila [148]

Answer:

D

Step-by-step explanation:

i think so but it will help you

8 0
3 years ago
PLEASE HELP, I WILL UP POINTS AND VOTE BRAINLIEST! DONT MAKE ME WASTE MY 88 POINTS.
Allushta [10]

Answer:

Ok im  using a standered form calculator:

1. 9/7x + 1/7 = -8

2. y = -6/5x - 3 =  -11

3. y = 1/3x - 2 = -10

Step-by-step explanation:

I will be glad to help more

6 0
3 years ago
URGENT: If θ is a second-quadrant angle and cosθ = -2/3, then tanθ = _____.
dangina [55]

In the second quadrant, both cos and tan are negative while only sin is positive.

To find tan, we will use the following property below:

\large \boxed{ {tan}^{2}  \theta  =  {sec}^{2}   \theta - 1}

Sec is the reciprocal of cos. If cos is a/b then sec is b/a. Since cos is 2/3 then sec is 3/2

\large{ {tan}^{2}  \theta =  {( -  \frac{3}{2}) }^{2}  - 1} \\   \large{ {tan}^{2}  \theta =   \frac{9}{4}  - 1} \\   \large{ {tan}^{2}  \theta = \frac{9}{4}   -  \frac{4}{4} \longrightarrow  \frac{5}{4}  } \\  \large{tan \theta =  \frac{ \sqrt{5} }{ \sqrt{4} } } \\  \large \boxed{tan \theta =  \frac{ \sqrt{5} }{2} }

Since tan is negative in the second quadrant. Hence,

\large{ \cancel{ tan \theta  =  \frac{ \sqrt{5} }{2} } \longrightarrow \boxed{tan \theta =  -  \frac{ \sqrt{5} }{2} }}

Answer

  • tan = -√5/2
3 0
3 years ago
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