Could I ask what you've used to make the animation repeat? From a quick search I noticed you aren't using setInterval() or d3.timer().
Are you just calling redraw as quickly as the CPU runs the code or am I missing something?
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Could I ask what you've used to make the animation repeat? From a quick search I noticed you aren't using setInterval() or d3.timer().
Are you just calling redraw as quickly as the CPU runs the code or am I missing something?
This is great, but I have a slight complaint. Sliding the kurtosis indicator changes the left distribution, which makes sense. However, it also changes the appearance of the right, normal distribution, which is misleading. Normal distributions have [EDIT: constant] kurtosis. I realize that the appearance is changing because the scale is changing so that the max is always pegged. However, it might be less confusing if…
The scales on the walks should be synchronized to the extremes across both images. I was trying to compare the two walks visually and realized that the scaled didn't match.
I worked with an exponential distribution before at work. At first, it seemed like we could model them like gaussian because the part of the data we were interested in was "close enough" to a gaussian. We already wrote code that worked for our other data that was gaussian. As it turns out, I was wrong. The thing that can't be easily seen in pictures is that exponential distributions move differently than gaussian dis…
This is great, but I have a slight complaint. Sliding the kurtosis indicator changes the left distribution, which makes sense. However, it also changes the appearance of the right, normal distribution, which is misleading. Normal distributions have [EDIT: constant] kurtosis. I realize that the appearance is changing because the scale is changing so that the max is always pegged. However, it might be less confusing if…
Agree: This is great, slight complaint... The scales on the walks should be synchronized to the extremes across both images. I was trying to compare the two walks visually and realized that the scaled didn't match.
The way the scales adjust on-the-fly means that once each graph (inevitably) either hits the near-top or near-bottom of the graph, the line stays there, and doesn't appear to move as the scale adjusts around it - the scale adjusts to keep the line close to the "adjustment point" near the edge of the graph.
There is a term for dealing with actual distributions that differ from the one you put as basis in your theory. http://en.wikipedia.org/wiki/Robust_statistics
Regression analysis and particulary ANOVA are sensitive to http://en.wikipedia.org/wiki/Heteroscedasticity but may be less sensitive to fat tails as long as the standard deviation is independent from the mean.
Earlier quoted context omitted.
Symmetric random walks, including those with step sizes drawn from a Gaussian process with mean zero, have expectation 0 at any time t. Since there's no drift term, either walk won't be expected to slowly deviate from zero.
This is false. Think about it this way. A random walk is a martingale - you're expectation of where you'll be in the future is. So you're right that, before you start the walk, they have expectation of being at 0 at any time in the future. But the variance of your probability distribution of where you'll be at time t is linear in t . So say that your variance is v(t) = t . Then at t =1, there is a 32% chance that you…
Is this a typo, or is teacher.js a cute way of saying JavaScript teacher?
Earlier quoted context omitted.
Symmetric random walks, including those with step sizes drawn from a Gaussian process with mean zero, have expectation 0 at any time t. Since there's no drift term, either walk won't be expected to slowly deviate from zero.
This is false. Think about it this way. A random walk is a martingale - you're expectation of where you'll be in the future is. So you're right that, before you start the walk, they have expectation of being at 0 at any time in the future. But the variance of your probability distribution of where you'll be at time t is linear in t . So say that your variance is v(t) = t . Then at t =1, there is a 32% chance that you…