Let X1,X2,… be a Bernoulli process. We will define some new sequences of random variables and inquire whether they form a Bernoulli process.
1. Let Yn = X2n. Is the sequence Yn a Bernoulli process?
2. Let Un = Xn+1. Is the sequence Un a Bernoulli process?
3. Let Vn = Xn+Xn+1. Is the sequence Vn a Bernoulli process?
4. Let Wn = (−1)nXn. Is the sequence Wn a Bernoulli process?
1. Yes
2. Yes
3. No
4. No
1. "Well, well, well, if it isn't the mysterious sequence Yn = X2n. Is Yn a Bernoulli process? Let's see. In a Bernoulli process, each random variable is independent and takes on only two possible values, usually 0 and 1. Now, in this case, since Yn is defined as X2n, it means we're skipping every other X value. Hmmm, sounds like Yn might be up to something sneaky. It's definitely not a straightforward Bernoulli process. Yn might be trying to throw a curveball at us! Sneaky, sneaky Yn!"
2. "Ah, the sequence Un = Xn+1. Is Un a Bernoulli process? Let's analyze it together, shall we? In a Bernoulli process, each random variable is independent and takes on two possible values. Now, in this case, we're shifting the indices by 1. It's like playing a little game of 'Where's Xn?' But wait, Un is just the next X value in line! It's not trying to trick us or hide anything. So, my friend, Un is indeed a Bernoulli process. No shenanigans here, just good old-fashioned randomness."
3. "Now we have the sequence Vn = Xn+Xn+1. What is Vn trying to accomplish? Is it trying to be a Bernoulli process? Let's find out. In a Bernoulli process, each random variable is independent and takes on two possible values. But in this case, Vn is formed by summing two consecutive X values. It's like a dynamic duo of randomness! So, Vn is different from the usual Bernoulli process. It's like a special edition or a remix of randomness. It's Vn's way of spicing things up, I suppose!"
4. "Ohoho, the mysterious Wn = (-1)nXn. What kind of mischief is Wn trying to pull? Is it a Bernoulli process in disguise? Let's investigate. In a Bernoulli process, each random variable is independent and takes on two possible values. Now, in this case, Wn seems to be flipping the sign of Xn based on whether n is even or odd. It's like Xn has been caught in a time-loop! So, Wn is definitely not your typical Bernoulli process. It's like a prankster, playing tricks with signs and numbers. Oh, Wn, you do love to keep us on our toes!"
To determine whether each of the given sequences form a Bernoulli process, we need to check if they satisfy the four properties of a Bernoulli process:
1. The random variables are independent: Each random variable in the sequence should be independent of the other random variables.
2. Each random variable has two possible outcomes: Each random variable in the sequence should have two possible outcomes, typically denoted as success and failure.
3. The probability of success is constant: The probability of success for each random variable in the sequence should be the same.
4. The trials are identical: Each random variable in the sequence should have the same distribution.
Let's analyze each sequence one-by-one:
1. Yn = X2n:
- The random variables Yn are not independent since they are taking values from every other term of Xn. In a Bernoulli process, each random variable must be independent.
- Therefore, Yn does not form a Bernoulli process.
2. Un = Xn+1:
- The random variables Un are dependent on the previous term of Xn, so they are not independent.
- Therefore, Un does not form a Bernoulli process.
3. Vn = Xn + Xn+1:
- The random variables Vn are dependent on both Xn and Xn+1, so they are not independent.
- Therefore, Vn does not form a Bernoulli process.
4. Wn = (-1)^n * Xn:
- The random variables Wn alternate in sign based on the value of n. This violates the condition that the probability of success should be constant for each random variable in a Bernoulli process.
- Therefore, Wn does not form a Bernoulli process.
Based on the analysis, none of the sequences Yn, Un, Vn, or Wn form a Bernoulli process.
To determine whether each of the new sequences Yn, Un, Vn, and Wn form a Bernoulli process, we need to check if they satisfy the three defining properties of a Bernoulli process: independence, identical distribution, and memorylessness. Let's analyze each sequence:
1. Yn = X2n:
To check if Yn forms a Bernoulli process, we need to verify the three properties.
- Independence: For any pair of indices i and j, Yi and Yj need to be independent. In this case, Yn = X2n, so to determine the independence, we need to examine the independence of X2n. Since X1, X2, ... are given to be independent, Yn = X2n will also be independent, making it a Bernoulli process.
- Identical distribution: For all indices i, Yi should have the same distribution. In this case, since Yn = X2n, they have the same distribution as Xn. Therefore, Yn has an identical distribution, satisfying the requirement for a Bernoulli process.
- Memorylessness: For all indices i, the probability of Yi+1 = 1 should be the same regardless of the outcomes of Y1, Y2, ..., Yi. In this case, Yn = X2n, and the outcome of X2n+1 is determined only by X2n. Therefore, Yn does not satisfy memorylessness, and it is not a Bernoulli process.
2. Un = Xn+1:
To analyze whether Un forms a Bernoulli process, we need to examine the three properties.
- Independence: For any pair of indices i and j, Ui and Uj need to be independent. Since X1, X2, ... are given to be independent, Un = Xn+1 will also be independent, making it a Bernoulli process.
- Identical distribution: For all indices i, Ui should have the same distribution. In this case, since Ui = Xi+1, they have the same distribution as Xn. Therefore, Un has an identical distribution, satisfying the requirement for a Bernoulli process.
- Memorylessness: For all indices i, the probability of Ui+1 = 1 should be the same regardless of the outcomes of U1, U2, ..., Ui. In this case, Ui+1 = Xi+2 is only determined by Xi+1. Therefore, Un satisfies memorylessness, and it is a Bernoulli process.
3. Vn = Xn+Xn+1:
To assess whether Vn forms a Bernoulli process, we need to evaluate the three properties.
- Independence: For any pair of indices i and j, Vi and Vj need to be independent. In this case, Vi = Xi+Xi+1, and the independence of Xi and Xi+1 is not given. Without knowing the independence relation between Xi and Xi+1, we cannot conclude if Vn satisfies independence, so it may or may not be a Bernoulli process.
- Identical distribution: For all indices i, Vi should have the same distribution. In this case, Vi = Xi+Xi+1, and they do not necessarily have the same distribution as Xi. Therefore, Vn does not satisfy identical distribution, and it is not a Bernoulli process.
- Memorylessness: For all indices i, the probability of Vi+1 = 1 should be the same regardless of the outcomes of V1, V2, ..., Vi. In this case, Vi+1 = Xi+1+Xi+2, and it depends on both Xi+1 and Xi+2. Therefore, Vn does not satisfy memorylessness, and it is not a Bernoulli process.
4. Wn = (-1)^n * Xn:
To determine if Wn forms a Bernoulli process, we need to evaluate the three properties.
- Independence: For any pair of indices i and j, Wi and Wj need to be independent. In this case, Wi = (-1)^i * Xi, and the dependence between (-1)^i and Xi is not known. Without knowing their independence relation, we cannot conclude if Wn satisfies independence, so it may or may not be a Bernoulli process.
- Identical distribution: For all indices i, Wi should have the same distribution. In this case, Wi = (-1)^i * Xi, and they do not have the same distribution as Xi. Therefore, Wn does not satisfy identical distribution, and it is not a Bernoulli process.
- Memorylessness: For all indices i, the probability of Wi+1 = 1 should be the same regardless of the outcomes of W1, W2, ..., Wi. In this case, Wi+1 = (-1)^(i+1) * Xi+1, and it depends on both (-1)^(i+1) and Xi+1. Therefore, Wn does not satisfy memorylessness, and it is not a Bernoulli process.
In conclusion:
- Yn is not a Bernoulli process.
- Un is a Bernoulli process.
- Vn is not a Bernoulli process.
- Wn is not a Bernoulli process.