analysis:course-w16:week10

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analysis:course-w16:week10 [2016/02/16 14:01] mvdm [Quantifying decoding accuracy] |
analysis:course-w16:week10 [2018/07/07 10:19] (current) |
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In general, from the [[http://en.wikipedia.org/wiki/Poisson_distribution | definition of the Poisson distribution]], it follows that | In general, from the [[http://en.wikipedia.org/wiki/Poisson_distribution | definition of the Poisson distribution]], it follows that | ||

- | \[P(n_i|\mathbf{x}) = \frac{(\tau f_i(\mathbf{x}))^{n_i}}{n_i!} e^{-\tau f_i (x)}\] | + | \[P(n_i|\mathbf{x}) = \frac{(\tau f_i(\mathbf{x}))^{n_i}}{n_i!} e^{-\tau f_i (\mathbf{x})}\] |

$f_i(\mathbf{x})$ is the average firing rate of neuron $i$ over $x$ (i.e. the tuning curve for position), $n_i$ is the number of spikes emitted by neuron $i$ in the current time window, and $\tau$ is the size of the time window used. Thus, $\tau f_i(\mathbf{x})$ is the mean number of spikes we expect from neuron $i$ in a window of size $\tau$; the Poisson distribution describes how likely it is that we observe the actual number of spikes $n_i$ given this expectation. | $f_i(\mathbf{x})$ is the average firing rate of neuron $i$ over $x$ (i.e. the tuning curve for position), $n_i$ is the number of spikes emitted by neuron $i$ in the current time window, and $\tau$ is the size of the time window used. Thus, $\tau f_i(\mathbf{x})$ is the mean number of spikes we expect from neuron $i$ in a window of size $\tau$; the Poisson distribution describes how likely it is that we observe the actual number of spikes $n_i$ given this expectation. | ||

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\[P(\mathbf{n}|\mathbf{x}) = \prod_{i = 1}^{N} \frac{(\tau f_i(\mathbf{x}))^{n_i}}{n_i!} | \[P(\mathbf{n}|\mathbf{x}) = \prod_{i = 1}^{N} \frac{(\tau f_i(\mathbf{x}))^{n_i}}{n_i!} | ||

- | e^{-\tau f_i (x)}\] | + | e^{-\tau f_i (\mathbf{x})}\] |

An analogy here is simply to ask: if the probability of a coin coming up heads is $0.5$, what is the probability of two coints, flipped simultaneously, coming up heads? If the coins are independent then this is simply $0.5*0.5$. | An analogy here is simply to ask: if the probability of a coin coming up heads is $0.5$, what is the probability of two coints, flipped simultaneously, coming up heads? If the coins are independent then this is simply $0.5*0.5$. |

analysis/course-w16/week10.1455649270.txt.gz ยท Last modified: 2018/07/07 10:19 (external edit)

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