question about probability - Page 2

ivanku · #11 04-18-2012, 12:59 AM

Quote:

Originally Posted by yaser

This is approximately ${1 \over e}$ since $\lim_{n\to\infty} (1-{1\over n})^n = {1\over e}$ . Numerically, ${1 \over e}\approx{1\over 2.718}\approx 0.37$ .

Is it correct to assume that $\nu_{min}$ from the coins tossing simulation in homework 2 assignment one should be close to that limit (where we're tossing 1000 coins 10 times and $\nu_{min}$ is the fraction of heads obtained for the coin which had the minimum frequency of heads)?

elkka · #12 04-18-2012, 03:49 AM

The law of big numbers states that the average $\nu_min$ is close to the $E{\nu_min}$.

$E\nu_min$ can be calculated directly for this experiment.
$P(\nu_min=0)$=0.623576
$P(\nu_min = 0.1)$ = 0.3764034
$P(\nu_min = 0.2)$ = 0.00002;
and $P(\nu_min>=0.3)=0$ for the purposes of calculating the mean.

Therefore, $E(\nu_min)$=0.037644, and the average proportion of heads for c_min should be close to this number.

SamK52 · #13 04-18-2012, 06:43 AM

Allow me to format your post:

Quote:

The law of big numbers states that the average $\nu_{min}$ is close to the $E{\nu_{min}}$ .

$E\nu_{min}$ can be calculated directly for this experiment.
$P(\nu_{min}=0)=0.623576$
$P(\nu_{min} = 0.1)= 0.3764034$
$P(\nu_{min} = 0.2) = 0.00002$
and $P(\nu_{min}>=0.3)=0$ for the purposes of calculating the mean.

Therefore, $E(\nu_{min})=0.037644$ , and the average proportion of heads for $c_{min}$ should be close to this number.

elkka · #14 04-18-2012, 11:26 AM

Quote:

Originally Posted by SamK52

Allow me to format your post:

Please, allow me to ask how you did it?

yaser · #15 04-18-2012, 01:57 PM

Quote:

Originally Posted by elkka

Please, allow me to ask how you did it?

http://book.caltech.edu/bookforum/sh...77&postcount=1

rohanag · #16 04-18-2012, 01:58 PM

how did you calculate those probability values? ( nu_min = 0, 0.1, 0.2 )

elkka · #17 04-19-2012, 04:10 AM

Thank you, Professor.

This how I calculate the probabilities. Let $h_{min} = 10*\nu_{min}$ - the number of heads for $c_{min}$ , and let

be the number of heads in i-th experiment (out of 1000). Then, as Professor has shown previously,

$P(\nu_{min}=0) =P(h_{min}=0) = P(\exists i\in[1,1000]\,\, h_i = 0)=1-P(\forall i\in[1,1000]\, h_i > 0)$
$=1-P(h_1 > 0)^{1000};$

Now, $P(h_1>0) = 1-P(h_1 = 0) = 1-0.5^{10}$ . Therefore,
$P(\nu_{min}=0)=1-(1-0.5^{10})^{1000} = 0.623576.$

Next,
$P(\nu_min=0.1)=P(h_{min}=1) =P(\forall i\in[1,1000] \,\,h_i > 0)-P(\forall i\in[1,1000] \,\,h_i > 1)$
$= P^{1000}(h_1>0)-P^{1000}(h_1>1) = (1-P(h_1=0))^{1000}-(1-P(h_1=0)-P(h_1=1))^{1000}$
$=(1-0.5^{10})^{1000}-(1-0.5^{10}-C_{10}^1*0.5^{10})^{1000} =0.3764034.$

Next,
$P(\nu_min=0.2)=P(h_{min}=2) =P(\forall i\in[1,1000]\,\, h_i > 1)-P(\forall i\in[1,1000] \,\,h_i > 2)$
$= P^{1000}(h_1>1)-P^{1000}(h_1>2)$
$= (1-P(h_1=0)-P(h_1=1))^{1000}-(1-P(h_1=0)-P(h_1=1)-P(h_1=2))^{1000}$
$=(1-0.5^{10}-C_{10}^1*0.5^{10})^{1000}-(1-0.5^{10}-C_{10}^1*0.5^{10}-C_{10}^2*0.5^{10})^{1000} =0.000204.$

The rest can be calculated directly too, but they are essenctially 0 for the purpose of calculating the mean.

rohanag · #18 04-19-2012, 08:44 AM

thank you for the detailed explanation.

nyxee · #19 07-13-2012, 05:55 PM

the answer is very clear but how do we know when to use this not.

to clarify my question, if say P(ten heads)=p and P(not ten heads)=q (=1-p). why does using (p^1000) give the wrong answer?

htlin · #20 07-15-2012, 04:56 AM

Quote:

Originally Posted by nyxee

the answer is very clear but how do we know when to use this not.

to clarify my question, if say P(ten heads)=p and P(not ten heads)=q (=1-p). why does using (p^1000) give the wrong answer?

$p^{1000}$ means the probability of getting ten heads in each of the independent random trials. Is that the event you are interested in?

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