User:DVO
For Prof. Welch
(This is also for anyone else who see this and happens to know the answer.)
Let's consider the continuous model for the Kalman filter learning tool's "sloshing" case, where the water is sloshing but not filling. Let be the height of the water at time t. Then
for some constants and , perhaps and .
In the continuous model, the "state vector" at time is .
(Eqn.1)
where is the (random) "process error."
We need to discretize this continuous system of ODEs.
For any and ,
is a fundamental matrix for , and . Any solution of Eqn. 1 can be written (by variation of parameters) as
.
Letting and , we have that
.
Thus the "discrete time state transition matrix" is
$A_{k-1} = \Phi(t_k, t_{k-1}) =
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
.
Is this a mistake in the "Team 18" document, or am I making some mistake?
This ends my question for Prof. Welch. Thanks.
Personal info
Name: Daniel O'Connor
(full name: Daniel Verity O'Connor)
Location: Los Angeles
Contributions
- Created Chain Rule article
- Created Fundamental Theorem of Calculus article
Random Math Problems
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.
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 comes up on the die?
The probability that it will take one roll is .
The probability that it will take two rolls is .
The probability that it will take three rolls is .
The probability that it will take four rolls is .
And so on.
So, the expected number of rolls that it will take to get a is:
.
What's the sum of this infinite series? It looks kind of like an infinite geometric series, but not exactly. Factoring out a makes it look a bit more like an infinite geometric series:
This is similar to a geometric series, which we know how to sum. But we have those annoying factors of , , , 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:
.
Differentiating both sides of this with respect to , we find:
.
So, we find that
.
Which seems like a very reasonable answer, given that the die has six sides.
What's the expected number of rolls it will take in order to get all six numbers at least once?
As a first step, let's find out the expected number of rolls it will take in order to get both and at least once.
Let be an integer. What's the probability that it will take rolls to get a and a ?
Consider two different events: Event is the event that the roll is a , and the first rolls give at least one and also no 's. Event is the event that the roll is a , and the first rolls give at least one and also no 's. The probability that it will take rolls to get a and a is .
What is ?
Let be the event that the roll is a . Let be the event that the first rolls give at least one . Let be the event that the first rolls give no 's.
.
Event is independent of the event , so:
.
is , but what is ?
Events and aren't independent. But we can say this:
. (Here is the complement of the event . I have used the fact that .)
.
.
Thus, $P(B \cap C)= \left(\frac56 \right)^{n-1} \left(1- \frac{ (\frac46)^{n-1} }{ (\frac56)^{n-1} } \right)
= \left(\frac56 \right)^{n-1} - \left(\frac46 \right)^{n-1}$ (Error compiling LaTeX. Unknown error_msg).
So .
We have found . We're close to finding the probability that it will take n rolls to get both a and a .
. Thus the probability that it will take rolls to get both a and a is .
Okay.....
So what's the expected number of rolls it will take to get both a and a ?
It is:
.
The expected number of rolls is .
Using a similar (but a little more complicated) method, I get that the expected number of rolls until all six numbers appear is .
I also found that the expected number of rolls until , , and appear is .