Timeline for Help with understanding a normal distribution/probability question
Current License: CC BY-SA 3.0
14 events
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| Jul 14, 2013 at 17:46 | comment | added | user997112 | This is the bit I dont get (I get everything else): "Let u denote the value sampled from the uniform distribution". What exactly is u here? I understand you've got <this value> and you're now looking at the uniform distribution (graph?) to see which of the 3 areas it lies in, but where does u come from and what type of value is it? Thanks for the ongoing help | |
| Jul 14, 2013 at 17:40 | comment | added | QuantIbex | That is exactly what is explained in the answer. I can't explain it in more basic terms. It seems that I won't be able to help you further on this. | |
| Jul 14, 2013 at 17:24 | comment | added | user997112 | Ok- they didn't say "plot" but thats what I did to try and understand what was happening. Could you say in basic terms why we are using the uniform distribution and how exactly does the uniform distribution, along with the probabilities calculated previously, determine which event is most likely to happen next? Thanks | |
| Jul 14, 2013 at 8:22 | history | edited | QuantIbex | CC BY-SA 3.0 |
added 49 characters in body
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| Jul 14, 2013 at 7:52 | comment | added | QuantIbex | I don't see were they say to plot the uniform distribution. The "according to the above probabilities to determine which order arrived first" is what I explained in the last two sentences of my answer. See edit for an example. | |
| Jul 14, 2013 at 7:51 | history | edited | QuantIbex | CC BY-SA 3.0 |
Added example
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| Jul 13, 2013 at 17:56 | comment | added | user997112 | This is the part I don't get- the next part of the paper says to plot the uniform distribution and "according to the above probabilities to determine which order arrived first". The above probabilities are going to be proportional to lambda_i. I dont understand how plotting them on the uniform distribution would give the next book state. If you could shed any light on this, it would be most appreciated. | |
| Jul 13, 2013 at 17:54 | comment | added | QuantIbex | Yes, within one iteration $\sum \lambda_k$ is constant, and the $k$ with the largest $\lambda_k$ will also have the largest $p_k$. But what do you infer from that? | |
| Jul 13, 2013 at 17:21 | comment | added | user997112 | Yes I didn't mean that part. I meant when we calculate each X_k on the first run, the equation is lambda_k divided by the sum of lambda_i. Now the sum of lambda_i will be the same for every calculation of lambda_k.... because the set doesn't change whilst we do this calculation. So for every calculation of lambda_k, it will be lambda_i divided by the same sum each time. As we are dividing through by the same sum every time, surely this is pointless and the largest probability will just be the largest lambda_i? | |
| Jul 13, 2013 at 17:16 | comment | added | QuantIbex | From the article: "After the next state of the order book is computed we recompute the $\lambda_i$ ’s since they depend on the order book". | |
| Jul 13, 2013 at 17:10 | comment | added | user997112 | No what I mean is, when calculating each X_k you divide lambda_k by the sum of all lambda_i. The sum of all lambda_i value will be the same for each calculation of X_k, surely? | |
| Jul 13, 2013 at 17:08 | comment | added | QuantIbex | I don't get it. If one $\lambda_k$ changes, then $\sum \lambda_k$ will change. For the simulation, if you pick the $k$ having the largest $\lambda_k$, say $\tilde{k}$, then you pick $\tilde{k}$ with probability one, which is not equal to $p_{\tilde{k}}$. | |
| Jul 13, 2013 at 16:17 | comment | added | user997112 | Thanks for this. My question is thus- the sum of lambda_i will not change for each lambda_k we calculate. As we also have a uniform distribution- so wouldn't the largest probability simply be the largest lambda_k? | |
| Jul 13, 2013 at 16:10 | history | answered | QuantIbex | CC BY-SA 3.0 |