Let N,X1,Y1,X2,Y2,… be independent random variables. The random variable N takes positive integer values and has mean a and variance r. The random variables Xi are independent and identically distributed with mean b and variance s, and the random variables Yi are independent and identically distributed with mean c and variance t. Let
A=∑i=1NXi and B=∑i=1NYi.
Find cov(A,B). Express your answer in terms of the given means and variances using standard notation.
cov(A,B)=?
Find var(A+B). Express your answer in terms of the given means and variances using standard notation.
var(A+B)=?
please someone 2 hours left
1 hour don't torture me
please help us . we need help anyone out there ?
cov(A,B) = b*c*r
var(A+B) = a*(s+t)+r*(c+b)^2
For var(a+b), you need to include 2 x the covariance
why the convariance need to be multiplied by 2? can anyone elucidate this part?
Because the variance of the sum of two RV's is only equal to the sum of the variances if the two RV's are independent. If the two RV's are dependent then the variance of the sum of the two RV's = sum of the variances + 2*COV(the two RV's)
cov(A,B) = b*c*r
var(A+B) = a*(s+t)+r*(c^2+b^2)+2*b*c*r
To find cov(A, B), we need to know the covariance between Xi and Yi for any i.
The covariance between two random variables X and Y can be defined as:
cov(X, Y) = E[(X - E[X])(Y - E[Y])]
where E[X] and E[Y] are the means of X and Y, respectively.
In this case, we are given that E[Xi] = b and E[Yi] = c for any i.
Knowing this, let's calculate cov(A, B):
cov(A, B) = cov(∑i=1NXi, ∑i=1NYi)
Since Xi and Yi are independent for any i, the covariance between the sums of independent variables is zero. Therefore:
cov(A, B) = ∑i=1N cov(Xi, Yi)
Since cov(Xi, Yi) = 0 for any i, we have:
cov(A, B) = ∑i=1N 0 = 0
Therefore, cov(A, B) = 0.
Now, let's find var(A+B):
var(A+B) = var(A) + var(B) + 2 * cov(A, B)
We already found that cov(A, B) = 0, so the equation simplifies to:
var(A+B) = var(A) + var(B)
To find var(A), we need to know the variance of A, which can be calculated as follows:
var(A) = var(∑i=1NXi)
Since Xi are independent and identically distributed, we can use the property that the variance of the sum of independent variables is the sum of their variances:
var(A) = ∑i=1N var(Xi)
Given that var(Xi) = s for any i, we have:
var(A) = N * s
Similarly, for var(B):
var(B) = N * t
Therefore, var(A+B) = var(A) + var(B) = N * s + N * t = N * (s + t).
So, var(A+B) = N * (s + t).
In summary:
cov(A, B) = 0
var(A+B) = N * (s + t)