For each of the following sequences, determine the value to which it converges in probability.
(a) Let X1,X2,… be independent continuous random variables, each uniformly distributed between −1 and 1.
Let Ui=X1+X2+⋯+Xii,i=1,2,….
What value does the sequence Ui converge to in probability?
0 - correct
Let Wi=max(X1,X2,…,Xi),i=1,2,….
What value does the sequence Wi converge to in probability?
1 - correct
Let Vi=X1⋅X2⋅⋯⋅Xi,i=1,2,….
What value does the sequence Vi converge to in probability?
0 - correct
(b) Let X1,X2,… be independent identically distributed random variables with E[Xi]=2 and var(Xi)=9, and let Yi=Xi/2i.
What value does the sequence Yi converge to in probability?
0 - correct
Let An=1n∑i=1nYi. What value does the sequence An converge to in probability?
0 - correct
Let Zi=13Xi+23Xi+1 for i=1,2,…, and let Mn=1n∑i=1nZi for n=1,2,….
What value does the sequence Mn converge to in probability?
Let Zi=13Xi+23Xi+1 for i=1,2,…, and let Mn=1n∑i=1nZi for n=1,2,….
What value does the sequence Mn converge to in probability?
Ans: 2
-1 - incorrect
Just kidding! Let's find out the correct answer with a touch of humor.
Well, to determine the value to which the sequence Mn converges in probability, we need to do a little bit of math (or clownery, if you will).
Since Zi = (1/3)Xi + (2/3)Xi+1, we can rearrange it as Zi - (2/3)Xi+1 = (1/3)Xi.
Now, taking the average of the above equation for n terms, we get Mn = (1/n)(Z1 - (2/3)X2 + Z2 - (2/3)X3 + ... + Zn - (2/3)X(n+1)).
If we do some clown algebra and simplify the expression, we can see that Mn is equal to (1/3)X1 + (1/3)X2 + ... + (1/3)X(n+1).
And since the random variables Xi are all independent and identically distributed, their expected value is 2 (given in the problem).
Therefore, the expected value of Mn is (1/3)(2)(n+1) = (2/3)(n+1).
As n approaches infinity, the expected value of Mn approaches infinity as well.
So, drum roll, please... The sequence Mn converges to positive infinity in probability (with a sidekick of clownery).
To determine the value to which the sequence Mn converges in probability, we need to calculate the expected value and variance of Zi.
First, let's calculate the expected value of Zi:
E[Zi] = E[1/3 * Xi + 2/3 * Xi+1]
= 1/3 * E[Xi] + 2/3 * E[Xi+1]
= 1/3 * 2 + 2/3 * 2 (Using the fact that E[Xi] = 2 for all i)
= 2
Next, let's calculate the variance of Zi:
Var(Zi) = Var(1/3 * Xi + 2/3 * Xi+1)
= (1/3)^2 * Var(Xi) + (2/3)^2 * Var(Xi+1) + 2 * (1/3) * (2/3) * Cov(Xi, Xi+1)
= (1/9) * 9 + (4/9) * 9 + 0 (Using the fact that Var(Xi) = 9 for all i, and Cov(Xi, Xi+1) = 0 since Xi and Xi+1 are independent)
= 3 + 4
= 7
Now, we can use the Weak Law of Large Numbers (WLLN) to determine the value to which the sequence Mn converges in probability:
Mn = 1/n * (Z1 + Z2 + ... + Zn)
According to WLLN, if E[Zi] = 2 and Var(Zi) = 7 for all i, then the sequence Mn converges to E[Zi] = 2 in probability.
Therefore, the value to which the sequence Mn converges in probability is 2.
To determine the value to which the sequence Mn converges in probability, we need to calculate the expected value and variance of Zi and then use the weak law of large numbers.
First, let's calculate the expected value and variance of Zi:
E[Zi] = E[1/3(Xi) + 2/3(Xi+1)]
= 1/3 * E[Xi] + 2/3 * E[Xi+1]
= 1/3 * 2 + 2/3 * 2 (since E[Xi] = 2 from the given information)
= 2
Var(Zi) = Var(1/3(Xi) + 2/3(Xi+1))
= 1/9 * Var(Xi) + 4/9 * Var(Xi+1) (since X1, X2, ... are independent and identically distributed)
= 1/9 * 9 + 4/9 * 9 (since Var(Xi) = 9 from the given information)
= 9
Next, let's use the weak law of large numbers, which states that if X1, X2, ... are independent and identically distributed random variables with finite mean (in this case, 2) and finite variance (in this case, 9), then the sample mean will converge in probability to the population mean.
Therefore, the sequence Mn = 1/n * Σ(i=1 to n) Zi converges in probability to the expected value of Zi, which is 2.
So, the value to which the sequence Mn converges in probability is 2.