Convergence in probability.
For each of the following sequences, determine whether it converges in probability to a constant. If it does, enter the value of the limit. If it does not, enter the number “999".
1) Let X1, X2,… be independent continuous random variables, each uniformly distributed between −1 and 1.
a) Let Ui=X1+X2+⋯+Xii, i=1,2,…. What value does the sequence Ui converge to in probability? (If it does not converge, enter the number “999". Similarly in all below.)
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b) Let Σi=X1+X2+⋯+Xi, i=1,2,…. What value does the sequence Σi converge to in probability?
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c) Let Ii=1 if Xi≥1/2, and Ii=0, otherwise. Define,
Si=(I1+I2+⋯+Ii)/i.
What value does the sequence Si converge to, in probability?
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d) Let Wi=max{X1,…,Xi}, i=1,2,…. What value does the sequence Wi converge to in probability?
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e) Let Vi=X1⋅X2⋯Xi, i=1,2,…. What value does the sequence Vi converge to in probability?
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2) Let X1, X2,…, be independent identically distributed random variables with E[Xi]=2 and Var(Xi)=9, and let Yi=Xi/2i.
a) What value does the sequence Yi converge to in probability?
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b) Let An=1n∑i=1nYi. What value does the sequence An converge to in probability?
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c) 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?
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See Similar Questions below.
1) Let's go through each part of the question:
a) For the sequence Ui, each term is the sum of i independent random variables uniformly distributed between -1 and 1. As i increases, the sum of these random variables will tend to cluster around 0, since there is an equal chance of positive and negative values. Therefore, the sequence Ui converges in probability to 0.
b) For the sequence Σi, each term is the sum of i independent random variables uniformly distributed between -1 and 1. Similar to part a), as i increases, the sum will tend to cluster around 0. Therefore, the sequence Σi also converges in probability to 0.
c) For the sequence Si, each term is the average of i independent indicator random variables. These indicator random variables are 1 if the corresponding Xi is greater than or equal to 1/2, and 0 otherwise. As i increases, the average will tend to converge to the probability that Xi is greater than or equal to 1/2. Since Xi is uniformly distributed between -1 and 1, the probability that Xi is greater than or equal to 1/2 is 1/2. Therefore, the sequence Si converges in probability to 1/2.
d) For the sequence Wi, each term is the maximum of the first i random variables. As i increases, the maximum value will tend to approach the upper bound of the range, which is 1. Therefore, the sequence Wi converges in probability to 1.
e) For the sequence Vi, each term is the product of the first i random variables. As i increases, the product will tend to approach 0, since any random variable close to -1 or 1 will heavily influence the product. Therefore, the sequence Vi converges in probability to 0.
2) Let's go through each part of the question:
a) For the sequence Yi, each term is Xi divided by 2^i. As i increases, 2^i becomes larger and larger, effectively shrinking the values of Yi towards 0. Therefore, the sequence Yi converges in probability to 0.
b) For the sequence An, each term is the average of the first n Yi values. Since each Yi tends to 0 as mentioned in part a), the average will also tend to 0. Therefore, the sequence An converges in probability to 0.
c) For the sequence Mn, each term is the average of the first n Zi values. Zi is obtained by taking a linear combination of two consecutive Xi values. Since each Xi is independent and has a mean of 2, the expectation of Zi would be (1/3)(2) + (2/3)(2) = 2. Therefore, the sequence Mn converges in probability to 2.
So, for the given sequences, the convergence in probability is as follows:
a) Ui converges to 0.
b) Σi converges to 0.
c) Si converges to 1/2.
d) Wi converges to 1.
e) Vi converges to 0.
a) Yi converges to 0.
b) An converges to 0.
c) Mn converges to 2.
To determine whether a sequence converges in probability to a constant, the following steps can be followed:
1) Calculate the expected value (mean) of the sequence.
2) Calculate the variance of the sequence.
3) If the expected value converges to a constant as the sequence size increases, and the variance approaches zero as the sequence size increases, then the sequence converges in probability to the constant.
Let's go through each part of the question and determine the convergence for each sequence.
1)
a) For Ui = X1 + X2 + ... + Xi, each Xi is a uniformly distributed random variable between -1 and 1. The expected value of Xi is 0 and the variance is 1/3. The expected value of Ui is i * 0 = 0. As i increases, the variance of Ui will increase without bound. Therefore, the sequence Ui does not converge in probability to a constant. Enter "999".
b) For Σi = X1 + X2 + ... + Xi, the expected value of Xi is 0 and the variance is i/3. As i increases, the expected value of Σi will approach infinity. Therefore, the sequence Σi does not converge in probability to a constant. Enter "999".
c) For Ii = 1 if Xi ≥ 1/2, and Ii = 0 otherwise, the expected value of Ii is P(Xi ≥ 1/2) = 1/2 and the variance is (1/2)(1 - 1/2) = 1/4. As i increases, the expected value of Si will approach 1/2 and the variance will approach 0. Therefore, the sequence Si converges in probability to 1/2.
d) For Wi = max{X1, ..., Xi}, the expected value of Wi is the expected value of Xi, which is 0, and the variance is (1/3)i. As i increases, the expected value of Wi will approach 0 and the variance will approach 0. Therefore, the sequence Wi converges in probability to 0.
e) For Vi = X1 * X2 * ... * Xi, the expected value of Vi is the product of the expected values of Xi, which is 0, and the variance is the product of the variances of Xi, which is (1/3)i. As i increases, the expected value of Vi will approach 0 and the variance will approach 0. Therefore, the sequence Vi converges in probability to 0.
2)
a) For the sequence Yi = Xi / 2^i, the expected value of Yi is (1/2^i) * 2 = 2 / 2^i, which approaches 0 as i increases. Additionally, the variance of Yi is (1/2^i)^2 * Var(Xi) = (1/2^i)^2 * 9 = 9 / 4^i, which also approaches 0 as i increases. Therefore, the sequence Yi converges in probability to 0.
b) For the sequence An = (1/n) * Σi=1 to n Yi, the expected value of An is (1/n) * n * 2 / 2^n = 2 / 2^n, which approaches 0 as n increases. Additionally, the variance of An is (1/n^2) * n * (9 / 4^n) = 9 / (n * 4^n), which also approaches 0 as n increases. Therefore, the sequence An converges in probability to 0.
c) For Zi = (1/3)Xi + (2/3)Xi+1, the expected value of Zi is (1/3) * 2 + (2/3) * 2 = 2 and the variance is [(1/3)^2 * 9 + (2/3)^2 * 9 + 2*(1/3)*(2/3)*Cov(Xi, Xi+1)] = 6 / 9 = 2/3. Therefore, the sequence Mn = (1/n) * Σi=1 to n Zi will have an expected value of 2 and a variance that approaches 0 as n increases. Therefore, the sequence Mn converges in probability to 2.
In summary:
a) Ui: Does not converge (999)
b) Σi: Does not converge (999)
c) Si: Converges to 1/2
d) Wi: Converges to 0
e) Vi: Converges to 0
Yi: Converges to 0
An: Converges to 0
Mn: Converges to 2