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2 years ago

What is the solution to the system of equations? y=-5+3, y=1

Mathematics

1 answer:

andrew11 [14]2 years ago

7 0

Answer:

x = 2/5 and y =1 (2/5, 1)

Step-by-step explanation:

I will assume that you meant y = -5x + 3. If that's the case, then we equate

y = -5x + 3 to y = 1, obtaining:

-5x = -2, so that x = 2/5 and y =1.

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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:

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

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

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

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

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

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