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I am trying to get an estimate for the Drake Equation by simulation (context doesn't matter to this question)

I have reached a point where I have a number of planets, $n_e$, and 3 probabilities; $f_l, f_i, f_c$.

I want to take a sample from the distribution x~Bin($n_e, f_l$). Then use this to get a sample from this distribution y~Bin($x, f_i$). And finally get a sample from z~Bin($y, f_c$).

I know a shortcut is to simply take a sample from Bin~($n_e, f_l \cdot f_i \cdot f_c)$.

But I was wondering why you can take this shortcut (assuming the 3 probabilities are independent)? Why can you chain independent probabilities like this, instead of doing each one consecutively, and using the result for the next one? Is this something that can easily be proven?

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2 Answers 2

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Think of three row-vectors of $n_e$ cells, which can contain a 1 or a 0. I'll refer to these as our source rows.

In the first row cells have a $1$ with probability $f_l$;
in the second row cells have a $1$ with probability $f_i$;
in the third row cells have a $1$ with probability $f_c$,
and otherwise any cell will be $0$.

Now we add a fourth and fifth row, which are computed results under two schemes.

Under scheme 1, we cross out (ignore) the $i$th column of the result if the first row has a 0 in position $i$. Then in turn, if the second row has a 0, we cross out (ignore) those columns as well. We then copy the columns of the third row down to the result row, but only for the columns that are not crossed out. We the count how many $1$'s are in the scheme-1 result row (i.e. that are neither crossed out nor 0). This is the first scenario you described.

Under scheme 2, the $i$-th result cell will be 1 if all three source cells above it are 1 (this happens with probability $f_lf_if_c$). We then count the $1$'s in the scheme-2 result row. This is the shortcut you described.

Now the 1's in the scheme 1 result row are in exactly the same places as the 1's in the scheme 2 result row; the number of 1's in each case is therefore always the same.

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  • $\begingroup$ Thanks, that really helps visualise it. I appreciate it. Now I need to figure out if I know how to turn this into a proof! $\endgroup$ Commented Jun 8, 2016 at 13:25
  • $\begingroup$ Some things that may help you get started (if you haven't seen them already): The values in the cells are Bernoulli random variables. The row-and-column structure gives you subscripts. $\endgroup$
    – Glen_b
    Commented Jun 8, 2016 at 17:24
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Let $L=1$ if a planet develops life and zero otherwise, $I=1$ when the planet develops intelligent life and zero otherwise and $C=1$ when is releases signals and zero otherwise. Assume the the outcomes for $L,I,C$ are independent. (note that in case of independence it could be that for some outcomes $C=1$ while $L=0$ but then no signal will be sent because life didn't even develop).

Then, by definition $f_l=P(L=1)$, the probability that $L$ is one, and similar $f_i=P(I=1)$ and $f_c=P(C=1)$.

The probability that a planet sends a signal is then $P((L=1) \ \& \ (I=1) \ \& \ (C=1))$ (note that if $(L=0) \ \& \ (I=0) \ \& \ (C=1)$ then no signal can be sent because no life has developped).

$P((L=1) \ \ \& \ (I=1) \ \& \ (C=1))$ is, because of the independence assumption $P(L=1) P(I=1) P(C=1)=f_l f_i f_c$. So the probability that among the $n_e$ planets there are $x$ that send signals is given by the binomial density $Bin(n_e;f_l f_i f_c)$ i.e. $P(X=x)$ where $ X \sim Bin(n_e;f_l f_i f_c)$.

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  • $\begingroup$ Thanks a lot! That's exactly what I was trying to write down, but I wasn't quite sure how to word it in an understandable way. $\endgroup$ Commented Jun 8, 2016 at 14:38
  • $\begingroup$ Don't thank me but give the answer a vote if it helps you $\endgroup$
    – user83346
    Commented Jun 8, 2016 at 14:41

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