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@@ -612,6 +612,159 @@ var ptx_lunr_docs = [
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"title": "",
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"body": " Let indicate heads on the first and second flip, respectively, of a fair coin. Then: Joint Distribution for Indicator Random Variables 1\/4 1\/4 1\/4 1\/4 Then the marginal distributions are: Then, for any : So are independent. "
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},
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{
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"id": "notes-01-29",
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"level": "1",
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"url": "notes-01-29.html",
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"type": "Section",
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"number": "",
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"title": "Thursday, Jan 29",
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"body": " Thursday, Jan 29 This is an outline of the topics we covered in class. These notes are not a substitute for your own note-taking. I highly recommend that you take your own notes during class. If you ever miss a class for any reason, reach out to another student in class to get a copy of their notes. Expected Value Let be the roll of a fair D6. What is the average value of ? The sample space is . To find the average: That is: take each value of , multiply by the probability, and add all the results together. Given a distribution table: Distribution 1 0.1 2 0.1 3 0.1 4 0.1 5 0.1 6 0.5 Then the average is: Let be a discrete random variable taking values with probabilities . The expected value of is: Let indicate , with . Then: That is: \"E(indicator random variable) = Pr(event that it indicates)\". Suppose we flip a fair coin 100 times. Let indicate heads on each flip. Then , so for every . What if indicates a run of 4 heads starting at flip 3? That is, flips 3, 4, 5, and 6 must come up heads, and all other flips can come up either heads or tails. Since these flip results are independent, the probabilities multiply: Let be a continuous random variable with pdf . The expected value of is: It's worth putting this side-by-side with to compare the structure of each formula. These both say: \"multiply each value of the random variable by the probability, then accumulate all of those products\". Expected value is a weighted average of random variable values, with the probabilities as the weights. Let with pdf . Then: What would happen if you forgot the ? Then the calculation would become: This mistake will often be easy to catch. For example, this random variable takes values between 0 and 1, so it doesn't seem very likely that the average value is 1! Find given a distribution for . Distribution for -1 0.4 1 0.1 2 0.2 3 0.3 We can start by writing a distribution table for : Distribution for 1 0.5 4 0.2 9 0.3 Then: If takes on values with probabilities , and is any function, then: Revisiting the previous example, the theorem says we don't need to first create the distribution table for . We can use the distribution table for , and just apply the square to each value: The situation with continuous random variables is similar. Let with pdf . Then: Notice, in particular, that If are random variables with finite expected value and , then: Let be the number of heads in coin flips with bias . Then . Gross. Instead of calculating this directly, define to indicate heads on each flip. Then: Flip a fair coin 100 times. Then, the expected number of heads is: Flip a fair coin 100 times. What is the expected number of runs of 4 heads? As in the binomial EV calculation, define indicator random variables each indicating a run of 4 heads starting at the specified flip. Then for each . Let be the number of runs of 4 heads. Then: Consider a geometric distribution . Geometric Distribution with 1 2 3 4 Then is an infinite summation. Instead, consider the following argument. If we flip a coin until we see heads, we either see heads on flip 1 or not. In the first case, . In the second case, what is the average value of ? Starting at flip 2, it will take on average flips to see heads. Since we already flipped the coin once, the total number of flips will be . So, we can write: "
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},
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{
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"id": "notes-01-29-3-2",
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"level": "2",
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"url": "notes-01-29.html#notes-01-29-3-2",
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"type": "Example",
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"number": "70",
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"title": "",
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"body": " Let be the roll of a fair D6. What is the average value of ? The sample space is . To find the average: That is: take each value of , multiply by the probability, and add all the results together. "
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},
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{
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"id": "notes-01-29-3-3",
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"level": "2",
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"url": "notes-01-29.html#notes-01-29-3-3",
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"type": "Example",
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"number": "71",
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"title": "",
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"body": " Given a distribution table: Distribution 1 0.1 2 0.1 3 0.1 4 0.1 5 0.1 6 0.5 Then the average is: "
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},
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{
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"id": "def-discrete-EV",
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"level": "2",
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"url": "notes-01-29.html#def-discrete-EV",
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"type": "Definition",
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"number": "73",
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"title": "",
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"body": " Let be a discrete random variable taking values with probabilities . The expected value of is: "
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},
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{
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"id": "notes-01-29-3-5",
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"level": "2",
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"url": "notes-01-29.html#notes-01-29-3-5",
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"type": "Example",
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"number": "74",
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"title": "",
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"body": " Let indicate , with . Then: That is: \"E(indicator random variable) = Pr(event that it indicates)\". "
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},
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{
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"id": "notes-01-29-3-6",
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"level": "2",
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"url": "notes-01-29.html#notes-01-29-3-6",
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"type": "Example",
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"number": "75",
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"title": "",
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"body": " Suppose we flip a fair coin 100 times. Let indicate heads on each flip. Then , so for every . What if indicates a run of 4 heads starting at flip 3? That is, flips 3, 4, 5, and 6 must come up heads, and all other flips can come up either heads or tails. Since these flip results are independent, the probabilities multiply: "
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},
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{
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"id": "def-continuous-EV",
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"level": "2",
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"url": "notes-01-29.html#def-continuous-EV",
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"type": "Definition",
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"number": "76",
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"title": "",
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"body": " Let be a continuous random variable with pdf . The expected value of is: "
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},
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{
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"id": "notes-01-29-3-9",
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"level": "2",
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"url": "notes-01-29.html#notes-01-29-3-9",
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"type": "Example",
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"number": "77",
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"title": "",
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"body": " Let with pdf . Then: What would happen if you forgot the ? Then the calculation would become: This mistake will often be easy to catch. For example, this random variable takes values between 0 and 1, so it doesn't seem very likely that the average value is 1! "
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},
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{
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"id": "notes-01-29-3-10",
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"level": "2",
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"url": "notes-01-29.html#notes-01-29-3-10",
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"type": "Example",
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"number": "78",
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"title": "",
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"body": " Find given a distribution for . Distribution for -1 0.4 1 0.1 2 0.2 3 0.3 We can start by writing a distribution table for : Distribution for 1 0.5 4 0.2 9 0.3 Then: "
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},
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{
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"id": "notes-01-29-3-11",
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"level": "2",
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"url": "notes-01-29.html#notes-01-29-3-11",
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"type": "Theorem",
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"number": "81",
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"title": "",
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"body": " If takes on values with probabilities , and is any function, then: "
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},
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{
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"id": "notes-01-29-3-12",
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"level": "2",
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"url": "notes-01-29.html#notes-01-29-3-12",
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"type": "Example",
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"number": "82",
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"title": "",
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"body": " Revisiting the previous example, the theorem says we don't need to first create the distribution table for . We can use the distribution table for , and just apply the square to each value: "
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},
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{
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"id": "notes-01-29-3-13",
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"level": "2",
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"url": "notes-01-29.html#notes-01-29-3-13",
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"type": "Example",
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"number": "83",
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"title": "",
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"body": " The situation with continuous random variables is similar. Let with pdf . Then: Notice, in particular, that "
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},
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{
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"id": "thm-linearity",
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"level": "2",
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"url": "notes-01-29.html#thm-linearity",
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"type": "Theorem",
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"number": "84",
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"title": "",
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"body": " If are random variables with finite expected value and , then: "
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},
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{
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"id": "notes-01-29-3-15",
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"level": "2",
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"url": "notes-01-29.html#notes-01-29-3-15",
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"type": "Example",
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"number": "85",
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"title": "",
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"body": " Let be the number of heads in coin flips with bias . Then . Gross. Instead of calculating this directly, define to indicate heads on each flip. Then: "
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},
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{
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"id": "notes-01-29-3-16",
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"level": "2",
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"url": "notes-01-29.html#notes-01-29-3-16",
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"type": "Example",
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"number": "86",
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"title": "",
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"body": " Flip a fair coin 100 times. Then, the expected number of heads is: "
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},
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{
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"id": "notes-01-29-3-17",
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"level": "2",
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"url": "notes-01-29.html#notes-01-29-3-17",
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"type": "Example",
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"number": "87",
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"title": "",
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"body": " Flip a fair coin 100 times. What is the expected number of runs of 4 heads? As in the binomial EV calculation, define indicator random variables each indicating a run of 4 heads starting at the specified flip. Then for each . Let be the number of runs of 4 heads. Then: "
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},
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{
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"id": "notes-01-29-3-18",
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"level": "2",
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"url": "notes-01-29.html#notes-01-29-3-18",
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"type": "Example",
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"number": "88",
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"title": "",
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"body": " Consider a geometric distribution . Geometric Distribution with 1 2 3 4 Then is an infinite summation. Instead, consider the following argument. If we flip a coin until we see heads, we either see heads on flip 1 or not. In the first case, . In the second case, what is the average value of ? Starting at flip 2, it will take on average flips to see heads. Since we already flipped the coin once, the total number of flips will be . So, we can write: "
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},
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{
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"id": "quiz-01",
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"level": "1",
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@@ -698,7 +851,7 @@ var ptx_lunr_docs = [
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"level": "2",
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"url": "recitation-calculus-review.html#thm-FTC",
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"type": "Theorem",
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"number": "70",
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"number": "90",
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"title": "Fundamental Theorem of Calculus.",
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"body": " Fundamental Theorem of Calculus If is an antiderivative of i.e., if then: "
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},
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