(New page: I believe that by saying X=drand48() Y=drand48() because the drand48() function produces indep RVs you can say: D=X^2+Y^2 Just like in problem 4 of homework 6.) |
(Brian Thomas Rhea HW7 -- Apologies for my tardiness in getting this up!) |
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Just like in problem 4 of homework 6. | Just like in problem 4 of homework 6. | ||
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| + | Your suggestion isn't quite like problem 4 of homework 6. Recall that in the case of the problem cited, X and Y were ''Gaussian'' RV's. | ||
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| + | HW6.4 generated an exponential RV from two Gaussian RV's. So, our strategy should be to go in reverse -- that is, to generate a Gaussian RV from one or more exponential RV's. | ||
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| + | We can generate exponential RV's using the template established in HW7.1b: | ||
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| + | *Let <math>F(x)=1-e^{-\lambda x} (x \geq 0)</math> | ||
| + | *Then, as per HW7.1b, if we assign an RV <math>D=F^{-1} (\text{drand48()})</math>, we know that D will be ~exp(<math>\lambda</math>). | ||
| + | *The only thing left in this step, then, is to find <math>F^{-1}(x)</math> for any x, so that we know what number we generate for a given input. Since we know that <math>F(x)=1-e^{-\lambda x}</math>, it is easy to show that <math>F^{-1}(x)=\frac{-\ln(1-x)}{\lambda}</math> | ||
| + | *Thus, <math>D=\frac{-\ln(1-\text{drand48()})}{\lambda}</math> | ||
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| + | Does anybody have a suggestion on where to proceed from here? | ||
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| + | --[[User:Thomas34|Thomas34]] 20:08, 20 October 2008 (UTC) | ||
Latest revision as of 16:08, 20 October 2008
I believe that by saying
X=drand48() Y=drand48()
because the drand48() function produces indep RVs you can say:
D=X^2+Y^2
Just like in problem 4 of homework 6.
Your suggestion isn't quite like problem 4 of homework 6. Recall that in the case of the problem cited, X and Y were Gaussian RV's.
HW6.4 generated an exponential RV from two Gaussian RV's. So, our strategy should be to go in reverse -- that is, to generate a Gaussian RV from one or more exponential RV's.
We can generate exponential RV's using the template established in HW7.1b:
- Let $ F(x)=1-e^{-\lambda x} (x \geq 0) $
- Then, as per HW7.1b, if we assign an RV $ D=F^{-1} (\text{drand48()}) $, we know that D will be ~exp($ \lambda $).
- The only thing left in this step, then, is to find $ F^{-1}(x) $ for any x, so that we know what number we generate for a given input. Since we know that $ F(x)=1-e^{-\lambda x} $, it is easy to show that $ F^{-1}(x)=\frac{-\ln(1-x)}{\lambda} $
- Thus, $ D=\frac{-\ln(1-\text{drand48()})}{\lambda} $
Does anybody have a suggestion on where to proceed from here?
--Thomas34 20:08, 20 October 2008 (UTC)
