Question

Probability question - how did they come up with the probability mass function in the textbook?

Answer 1

See screenshot

Answer 2

\[\frac{1}{2^i}*\frac{1}{2^j}\]makes sense for i sons followed by j daughters, but I don't understand the last\[1/2^2\]

Answer 3

It's actually i+1 and j+1 that makes up for (i+j+2).
This also included those families where there are no children(i,e. i=0 and j=0)

Answer 4

Hmm why would it be i + 1? I guess the whole question is now confusing, I don't really follow why they chose that as the probability mass function

Answer 5

We can construct some pretty arbitrary PMF with the geometric series, as long as you know this simplification of the geometric series is true, we can take two arbitrary ones like this:

\[\frac{1}{1-x} = \sum_{i=0}^\infty x^i\]\[\frac{1}{1-y} = \sum_{j=0}^\infty y^j\]

Multiply them together:

\[\frac{1}{1-x} \frac{1}{1-y} = \sum_{i=0}^\infty x^i\ \sum_{j=0}^\infty y^j\]
Multiplication distributes over addition, try this out with like a set of like six numbers or something if you don't believe this and want to test out a smaller case of this:
\[\frac{1}{(1-x)(1-y)} = \sum_{i=0}^\infty \sum_{j=0}^\infty x^i\ y^j\]

We want the left hand side to be 1 in order to be normalized, so just multiply through by (1-x)(1-y)

\[1= \sum_{i=0}^\infty \sum_{j=0}^\infty x^i\ y^j(1-x)(1-y)\]

Their specific choice was:
\[x^i\ y^j(1-x)(1-y) =2^{-i-j-2}\]

Or just more simply since they chose the same for both:

\[x^i\ (1-x) =2^{-i-1}\]
Some funky algebra where I just show that corresponding parts are equal for solving for \(x=2^{-1}\)
\[x^i\ (1-x) =(2^{-1})^{i}(2^{-1}) = (2^{-1})^{i}(1-2^{-1}) \]

Hopefully that wasn't too crazy? I can clarify whatever though this was interesting.

Answer 6

Sorry, still wondering why\[\frac{1}{1-x}\]instead of just 1/x. It's been a while since I looked at math, and the textbook went from rolling a die to this

Answer 7

Ah ok that's true cause of this:

S is just this geometric series:
\[S=1+x+x^2+x^3+x^4+\cdots\]
Multiply both sides of the equation by x
\[xS=x+x^2+x^3+x^4+x^5\cdots\]
Add 1 to both sides

\[xS+1=1+x+x^2+x^3+x^4+x^5\cdots\]
Wait a sec, the right hand side is S!
\[xS+1=S\]
Subtract xS from both sides then factor it out:
\[1=(1-x)S\]

\[\frac{1}{1-x} = S\]

S is that infinite geometric series though, so

\[\frac{1}{1-x} = 1+x+x^2+x^3+x^4+\cdots\]