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1 year ago

Which relation is a function?

Mathematics

2 answers:

Artemon [7]1 year ago

8 0

Answer:

Step-by-step explanation:

What makes a function is if each input or x-value has EXACTLY one output or y-value.

a. is wrong because the x value 1, has both an output of 1 and 4

b. is wrong because -1 is repeated 4 times with all different outputs

c. is correct because each input has exactly one output. None of the x values repeat itself

d. is wrong because -1 has 2 outputs.

so this is what makes a function and what doesn't. Hope this helps!

Rainbow [258]1 year ago

6 0

Answer:

b is the answer so circle b if youneed anything else cantact me

Step-by-step explanation:

ok please follow me and support

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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

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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:

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

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