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| − | == For Prof. Welch ==
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| − | (This is also for anyone else who see this and happens to know the answer.)
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| − | Let's consider the continuous model for the Kalman filter learning tool's "sloshing" case, where the water is sloshing but not filling. Let <math>L(t)</math> be the height of the water at time t. Then
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| − | <math>L(t) = c + k_s \sin(\omega t)</math> for some constants <math>c</math> and <math>k_s</math>, perhaps <math>c=.3</math> and <math>k_s=.05</math>.
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| − | In the continuous model, the "state vector" at time <math>t</math> is <math>x(t) =
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| − | \begin{pmatrix}
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| − | L(t)\\
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| − | k_s \end{pmatrix}</math>.
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| − | <math>x'(t) = \begin{pmatrix}
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| − | 0& \omega \cos(\omega t)\\
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| − | 0&0 \end{pmatrix} x(t) + w(t)</math> (Eqn.1)
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| − | where <math>w(t)</math> is the (random) "process error."
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| − | We need to discretize this continuous system of ODEs.
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| − | For any <math>t</math> and <math>t_0</math>, <math>\Phi(t,t_0) = \begin{pmatrix}
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| − | 1&\sin(\omega t) - \sin(\omega t_0) \\
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| − | 0&1\end{pmatrix}</math>
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| − | is a fundamental matrix for <math>x'(t)=\begin{pmatrix}
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| − | 0& \omega \cos(\omega t)\\
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| − | 0&0 \end{pmatrix} x(t)</math>, and <math>\Phi(t_0,t_0)=\begin{pmatrix}
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| − | 1&0\\
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| − | 0&1\end{pmatrix}</math> . Any solution <math>x(t)</math> of Eqn. 1 can be written (by variation of parameters) as
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| − | <math>x(t) = \Phi(t,t_0)x(t_0) + \int_{t_0}^t \Phi(t,\tau)w(\tau) d \tau</math>.
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| − | Letting <math>t=t_k</math> and <math>t_0 = t_{k-1}</math> , we have that
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| − | <math>{x(t_k) = \Phi(t_k,t_{k-1})x(t_{k-1}) + \int_{t_{k-1}}^{t_k} \Phi(t_k,\tau)w(\tau) d \tau}</math>.
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| − | Thus the "discrete time state transition matrix" is
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| − | <math>A_{k-1} = \Phi(t_k, t_{k-1}) =
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| − | \begin{pmatrix}
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| − | 1&\sin(\omega t_k) - \sin(\omega t_{k-1}) \\
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| − | 0&1\end{pmatrix}</math> .
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| − | However, this disagrees with Equation (6) on page 4 of the document "Team 18: The Kalman Filter Learning Tool, Dynamic and Measurement Models," where it is stated that the discrete time state transition matrix is
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| − | <math>A(t) = \begin{pmatrix}
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| − | 1& \omega \cos(\omega t)\\
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| − | 0&1 \end{pmatrix}</math> .
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| − | Is this a mistake in the "Team 18" document, or am I making some mistake?
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| − | This ends my question for Prof. Welch. Thanks.
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| − | == Personal info ==
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| − | Name: Daniel O'Connor
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| − | (full name: Daniel Verity O'Connor)
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| − | Location: Los Angeles
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| − | == Contributions ==
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| − | * Created [[Chain Rule]] article
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| − | * Created [[Fundamental Theorem of Calculus]] article
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| − | == Random Math Problems ==
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| − | The following problems are part of a discussion I'm having with someone I know. Please don't comment about them, and most importantly please don't correct any errors or give any hints about better solutions.
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| − | Suppose you have a fair six-sided die and you're going to roll the die again and again indefinitely. What's the expected number of rolls until a <math>1</math> comes up on the die?
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| − | The probability that it will take one roll is <math>\frac{1}{6} </math>.
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| − | The probability that it will take two rolls is <math>\left(\frac56 \right)\left(\frac16 \right) </math>.
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| − | The probability that it will take three rolls is <math>\left(\frac56 \right)^2 \left(\frac16 \right) </math>.
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| − | The probability that it will take four rolls is <math>\left(\frac56 \right)^3 \left(\frac16 \right) </math>.
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| − | And so on.
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| − | So, the expected number of rolls that it will take to get a <math>1</math> is:
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| − | <math>1\cdot \frac{1}{6} + 2\cdot \left(\frac56 \right)\left(\frac16 \right) + 3\cdot \left(\frac56 \right)^2 \left(\frac16 \right) + 4 \cdot \left(\frac56 \right)^3 \left(\frac16 \right) + \cdots</math>.
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| − | What's the sum of this infinite series? It looks kind of like an infinite geometric series, but not exactly. Factoring out a <math>\frac16</math> makes it look a bit more like an infinite geometric series:
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| − | <math>\left(\frac16 \right) \left(1 + 2\cdot \left(\frac56 \right) + 3\cdot \left(\frac56 \right)^2 + 4 \cdot \left(\frac56 \right)^3 + \cdots \right)</math>
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| − | This is similar to a geometric series, which we know how to sum. But we have those annoying factors of <math>2</math>, <math>3</math>, <math>4</math>, etc. to worry about. Maybe we could play around with the formula for a geometric series to get a formula for this series. The formula for a geometric series is:
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| − | <math>1 + r + r^2 + r^3 + r^4 + \cdots = \frac{1}{1-r}</math>.
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| − | Differentiating both sides of this with respect to <math>r</math>, we find:
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| − | <math>1 + 2r + 3r^2 + 4r^3 + \cdots = -(1-r)^{-2}(-1) = \frac{1}{(1-r)^2}</math>.
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| − | So, we find that
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| − | <math>\left(\frac16 \right) \left(1 + 2\cdot \left(\frac56 \right) + 3\cdot \left(\frac56 \right)^2 + 4 \cdot \left(\frac56 \right)^3 + \cdots \right) = \frac16 \frac{1}{(1-\frac56)^2} = \frac16 (36) = 6</math>.
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| − | Which seems like a very reasonable answer, given that the die has six sides.
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| − | What's the expected number of rolls it will take in order to get all six numbers at least once?
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| − | As a first step, let's find out the expected number of rolls it will take in order to get both <math>1</math> and <math>2</math> at least once.
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| − | Let <math>n \ge 2</math> be an integer. What's the probability that it will take <math>n</math> rolls to get a <math>1</math> and a <math>2</math>?
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| − | Consider two different events: Event <math>OneFirst</math> is the event that the <math>n^{th}</math> roll is a <math>2</math>, and the first <math>n-1</math> rolls give at least one <math>1</math> and also no <math>2</math>'s. Event <math>TwoFirst</math> is the event that the <math>n^{th}</math> roll is a <math>1</math>, and the first <math>n-1</math> rolls give at least one <math>2</math> and also no <math>1</math>'s. The probability that it will take <math>n</math> rolls to get a <math>1</math> and a <math>2</math> is <math>P(OneFirst)+ P(TwoFirst)</math>.
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| − | What is <math>P(OneFirst)</math>?
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| − | Let <math>A</math> be the event that the <math>n^{th}</math> roll is a <math>2</math>. Let <math>B</math> be the event that the first <math>n-1</math> rolls give at least one <math>1</math>. Let <math>C</math> be the event that the first <math>n-1</math> rolls give no <math>2</math>'s.
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| − | <math>P(OneFirst)=P(A \cap B \cap C)</math>.
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| − | Event <math>A</math> is independent of the event <math>B \cap C</math>, so:
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| − | <math>P(A \cap (B \cap C)) = P(A)P(B \cap C)</math>.
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| − | <math>P(A)</math> is <math>\frac16</math>, but what is <math>P(B \cap C)</math>?
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| − | Events <math>B</math> and <math>C</math> aren't independent. But we can say this:
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| − | <math>P(B \cap C)= P(C)P(B|C)= P(C)\left(1-P(B^c|C)\right)</math>. (Here <math>B^c</math> is the complement of the event <math>B</math>. I have used the fact that <math>P(X|Y)+P(X^c|Y)=1</math>.)
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| − | <math>P(C) = \left(\frac56 \right)^{n-1} </math>.
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| − | <math>P(B^c|C) = \frac{P(B^c \cap C)}{P(C)}= \frac{(\frac46)^{n-1}}{(\frac56)^{n-1}} </math>.
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| − | Thus, <math>P(B \cap C)=
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| − | \left(\frac56 \right)^{n-1} \left(1- \frac{ (\frac46)^{n-1} }{ (\frac56)^{n-1} } \right)
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| − | = \left(\frac56 \right)^{n-1} - \left(\frac46 \right)^{n-1} </math>.
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| − | So <math>P(A \cap (B \cap C) ) = \left( \frac16 \right) \left( \left(\frac56 \right)^{n-1} - \left(\frac46 \right)^{n-1} \right)</math>.
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| − | We have found <math>P(OneFirst)</math>. We're close to finding the probability that it will take n rolls to get both a <math>1</math> and a <math>2</math>.
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| − | <math>P(TwoFirst) = P(OneFirst)</math>. Thus the probability that it will take <math>n</math> rolls to get both a <math>1</math> and a <math>2</math> is <math>P(OneFirst)+ P(TwoFirst) = \left( \frac13 \right) \left( \left(\frac56 \right)^{n-1} - \left(\frac46 \right)^{n-1} \right)</math>.
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| − | Okay.....
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| − | So what's the expected number of rolls it will take to get both a <math>1</math> and a <math>2</math>?
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| − | It is:
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| − | <math>{2 \cdot \left( \frac13 \right) \left( \left(\frac56 \right)^1 - \left(\frac46 \right)^1 \right) + 3 \cdot \left( \frac13 \right) \left( \left(\frac56 \right)^2 - \left(\frac46 \right)^2 \right) + 4 \cdot \left( \frac13 \right) \left( \left(\frac56 \right)^3 - \left(\frac46 \right)^3 \right) + \cdots}</math>
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| − | <math> = \left( \frac13 \right) \left( 2 \left(\frac56 \right)^1 + 3 \left(\frac56 \right)^2 + 4 \left(\frac56 \right)^3 + \cdots \right) - \left( \frac13 \right) \left( 2\left(\frac46 \right)^1 + 3 \left(\frac46 \right)^2 + 4 \left(\frac46 \right)^3 + \cdots \right) </math>
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| − | <math>= \left( \frac13 \right) \left( -1 + 1 + 2 \left(\frac56 \right)^1 + 3 \left(\frac56 \right)^2 + 4 \left(\frac56 \right)^3 + \cdots \right) - \left( \frac13 \right) \left(-1 + 1 + 2\left(\frac46 \right)^1 + 3 \left(\frac46 \right)^2 + 4 \left(\frac46 \right)^3 + \cdots \right)</math>
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| − | <math>= \left( \frac13 \right) \left( -1 + \frac{1}{(1-\frac56)^2} \right) - \left( \frac13 \right) \left( -1 + \frac{1}{(1-\frac46)^2} \right) </math>
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| − | <math>= \left( \frac13 \right) ( -1 + 36 + 1 - 9 ) = \left( \frac13 \right) (27) = 9</math>.
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| − | The expected number of rolls is <math>9</math>.
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| − | Using a similar (but a little more complicated) method, I get that the expected number of rolls until all six numbers appear is <math>14.7</math>.
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| − | I also found that the expected number of rolls until <math>1</math>, <math>2</math>, and <math>3</math> appear is <math>11</math>.
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