Base×hight×width you are just doing 3/4×3/4×3/4 and what ever your answer is put a little 3 above it so it almost looks like an exponent. Pretend that ABC is the answer it would look like this format = ABCinches3 but the three is tiny
Answer:
(a)![E[X+Y]=E[X]+E[Y]](https://tex.z-dn.net/?f=E%5BX%2BY%5D%3DE%5BX%5D%2BE%5BY%5D)
(b)
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).](https://tex.z-dn.net/?f=E%5Bf%28X%2CY%29%5D%3D%5Csum_%7Bx%2Cy%7Df%28x%2Cy%29%5Ccdot%20P%28X%3Dx%2C%20Y%3Dy%29.)
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).](https://tex.z-dn.net/?f=E%5BX%2BY%5D%3D%5Csum_%7Bx%2Cy%7D%28x%2By%29P%28X%3Dx%2CY%3Dy%29%5C%5C%3D%5Csum_%7Bx%2Cy%7DxP%28X%3Dx%2CY%3Dy%29%2B%5Csum_%7Bx%2Cy%7DyP%28X%3Dx%2CY%3Dy%29.)
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].](https://tex.z-dn.net/?f=%5Csum_%7Bx%2Cy%7DxP%28X%3Dx%2CY%3Dy%29%5C%5C%3D%5Csum_%7Bx%7Dx%5Csum_%7By%7DP%28X%3Dx%2CY%3Dy%29%5C%5C%5Ctext%7BTaking%20Marginal%20distribution%20of%20x%7D%5C%5C%3D%5Csum_%7Bx%7DxP%28X%3Dx%29%3DE%5BX%5D.)
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].](https://tex.z-dn.net/?f=%5Csum_%7Bx%2Cy%7DyP%28X%3Dx%2CY%3Dy%29%5C%5C%3D%5Csum_%7By%7Dy%5Csum_%7Bx%7DP%28X%3Dx%2CY%3Dy%29%5C%5C%5Ctext%7BTaking%20Marginal%20distribution%20of%20y%7D%5C%5C%3D%5Csum_%7By%7DyP%28Y%3Dy%29%3DE%5BY%5D.)
Combining these two gives the formula:

Therefore:
![E[X+Y]=E[X]+E[Y] \text{ as required.}](https://tex.z-dn.net/?f=E%5BX%2BY%5D%3DE%5BX%5D%2BE%5BY%5D%20%5Ctext%7B%20%20as%20required.%7D)
(b)We want to show that if X and Y are independent random variables, then:

By definition of Variance, we have that:
![Var(X+Y)=E(X+Y-E[X+Y]^2)](https://tex.z-dn.net/?f=Var%28X%2BY%29%3DE%28X%2BY-E%5BX%2BY%5D%5E2%29)
![=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)](https://tex.z-dn.net/?f=%3DE%5B%28X-%5Cmu_X%20%20%2BY-%20%5Cmu_Y%29%5E2%5D%5C%5C%3DE%5B%28X-%5Cmu_X%29%5E2%20%20%2B%28Y-%20%5Cmu_Y%29%5E2%2B2%28X-%5Cmu_X%29%28Y-%20%5Cmu_Y%29%5D%5C%5C%24Since%20we%20have%20shown%20that%20expectation%20is%20linear%24%5C%5C%3DE%28X-%5Cmu_X%29%5E2%20%20%2BE%28Y-%20%5Cmu_Y%29%5E2%2B2E%28X-%5Cmu_X%29%28Y-%20%5Cmu_Y%29%5D%5C%5C%3DE%5B%28X-E%28X%29%5D%5E2%20%20%2BE%5BY-%20E%28Y%29%5D%5E2%2B2Cov%20%28X%2CY%29)
Since X and Y are independent, Cov(X,Y)=0

Therefore as required:

Answer:
Option D
Step-by-step explanation:
Given is a table which gives the order pair of x and y for a function f(x)
We have to find the local minimum of the function f(x)
We have from the table the values of different f(x)
Of all we find the least value is -15 and for this x value is -2 or +2
Hence f(x) has minimum at two points
(-2,-15) and (2,-15)
Out of 4 options given we find that (2,-15) appears in IV option
Hence option D is right answer