Understanding Binomial PMF Notation: Clearing Up Confusion?

[EmmaSaunders1]

Hi,

could someone possible make something clear for me - I have come across this notation for a binomial PMF formed from an underlying beurnolli distribution:

PS_{n}(\bar{p}n)\sim\sqrt{\frac{1}{2\pi n\bar{p}(1-\bar{p})}}exp
[n\varnothing(p,\bar{p}] ,\\

PS_{n}(\bar{p}n)=PMF-of-binomial-dist-from-underlying-binary-PMF\\
where,pz(1)=p>0,pz(0)=q>0,q=(1-p)

=I can understand that this is the binomial PMF with variance = npq, and the square root term is easy to understand. I don't understand the term raised to the exponential and how one can get from the beurnolli distribution to this binomial distribution, could someone please clarify

 

[mathman]

You need to fix your equation notation. Is it supposed to be Latex?

 

[EmmaSaunders1]

Hi - yeah I used a Latex online generator - the gif file visualizes clearly the notation

 

[Rasalhague]

This is how it looks in the attachment. To get LaTeX to display on a separate line, put "tex" in square brackets, [ ], before the LaTeX, and "/text" in square brackets after it. If you want to use LaTeX in the same line as other text, type "itex" in square brackets at the beginning instead of "tex", and "/itex" at the end instead of "/tex".

PS_{n}(\bar{p}n)\sim\sqrt{\frac{1}{2\pi n\bar{p}(1-\bar{p})}}exp<br /> [n\varnothing(p,\bar{p}],

PS_{n}(\bar{p}n)= PMF-of-binomial-dist-from-underlying-binary-PMF

where,pz(1)=p&gt;0,pz(0)=q&gt;0,q=(1-p)

 

[EmmaSaunders1]

Thats correct yes

 

[EmmaSaunders1]

I appologise - the last bracket sequence should read like p)]

 

[pmsrw3]

Are you sure you're describing this correctly? This doesn't look to me like it has anything to do with the relationship between binomial and Bernoulli distributions. It looks to me like a normal approximation for the binomial. Except that I have not a clue what n\varnothing(p,\bar{p}) represents. Also, what exactly are pz and \bar{p}? How can PS_{n}(\bar{p}n) be the PMF for a binomial distribution, when its argument (apparently) takes on non-integer values?

Can you give us some context?

 

[EmmaSaunders1]

Hi Thanks for taking a look - it is a normal approximation of a binomial distribution from a beurnolli process. I am just having a little difficulty understanding the notation and would appreciate if anyone could point me in the correct direction to understand this.

Attached is a more detailed explanation.

Thanks all

 

[pmsrw3]

OK, that makes more sense. You've flipped a biased coin n times -- that's the Bernoulli process. The random variable S_n is the number of heads that turned up, and it has a binomial distribution. \overline{p} = \frac{\text{number heads}}{n}, so that \overline{p}n is an integer, the number of heads, between 0 and n. (Your text restricts it to between 1 and n-1 because \phi(p,\overline{p}) blows up at 0 and n.) \mbox{P}_{S_n}(k) is the probability that S_n=k. You're not supposed to be able to figure out yet where the RHS of 1.23 came from. The proof of it should follow. This is NOT yet a normal approximation -- presumably they'll get to that further on in the chapter.

I don't understand ... how one can get from the beurnolli distribution to this binomial distribution, could someone please clarify

Well, that's simple. A Bernoulli process is just flipping a (possibly biased) coin over and over and counting up the number of heads. Now, you should know that, if you flip a coin n times and count the number of heads, it will have a binomial distribution.

 

[EmmaSaunders1]

Thanks for that - the exponential term is then just set such that the probability of a number of success in any give n-trials experimet is between 0 and 1