Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is possibly a follow-up question to this one:

different probability spaces for $P(X=k)=\binom{n}{k}p^k\big(1-p\big)^{ n-k}$?

Consider the two models in the title:

  • a fair coin being tossed $n$ times
  • $n$ fair coins being tossed once

and calculate the probability in each model that "head" appear(s) $k~ (0\leq k\leq n)$ times. Then one may come up with the same answer that $$ P(\text{"head" appear(s)} ~k~ \text{times}) = \binom{n}{k}p^k\big(1-p\big)^{n-k} $$

However, the first one can be regarded as a random process, where the underlying probability space is $\Omega = \{0,1\}$ ($1$ denotes "head" and $0$ for "tail") and the time set $T=\{1,2,\cdots,n\}$. While in the second one, the underlying probability space is $\Omega = \{0,1\}^n$.

Here are my questions:

  • How can I come up with the same formula with these two different points of view?
  • Are these two models essentially the same?
share|improve this question
4  
Yes, unless $n$ is very very large, in which case the coin might wear out! –  André Nicolas Oct 4 '11 at 2:30
2  
In the first case, you are assuming that there are $n$ independent trials of the experiment of tossing a fair coin once. In the second case, you are assuming that the $n$ experiments with $n$ different fair coins are independent experiments. In both cases, you end up with a sample space $\{0, 1\}^n$ and the probability measure on this sample space comes from the independence assumptions and is the same in both cases. So, you should get the same formula in both cases and the two models are essentially the same. –  Dilip Sarwate Oct 4 '11 at 13:43
1  
@AndréNicolas When $n$ is very large, the first model assumes that one has a large amount of time to waste tossing the one coin, and the second model assumes that one has a large amount of money ($n$ coins!) and many hands to toss them all simultaneously! :-) –  Dilip Sarwate Oct 4 '11 at 13:49

2 Answers 2

up vote 5 down vote accepted

The models are essentially the same. I think this automatically answers your first question as well.

You can see the two as trading a space dimension for a time dimension.

share|improve this answer

Both models are basically a way to put a probability in $\{0,1\}^n$.

Usually you will be given a probability distribution in $\{0,1\}$, and try to extend it to a probability in $\{0,1\}^n$, according to some extra assumption.

If one experiment (tossing a coin) does not influence the other (tossing it again, or tossing another coin), then, you will have the model you describe.

The point is that when you talk about a random process, usually you are allowing that the result of an experiment (toss a coin) might influence the result of the next (toss it again). Changing this condition, you might get a different probability distribution in $\{0,1\}^n$.

For example, it might be assumed that when the outcome is $1$, then the probabilities for the next outcome are flipped, that is $p$ becomes $1-p$. A more concrete example is the probability of a certain letter appearing in a text. After a consonant, it will be likely that the next letter will be a vogue. After a "p", we will not be likely to get a "x" or a "w".

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.