Solve the problem involving toss of a biased coin- Probability

[chwala]

Homework Statement

See attached

Relevant Equations

Probability -Expectation and variance

This is the problem;

My thinking on this is based on Von Neumann Strategy i.e
$e=pf+(1-p)((f+e)$ where $e$= Expected value, $p$= Probability and $f$ = number of tosses ...in our case $f=1$
$e=\frac{f}{p}=\frac{1}{p}$ This is clear (as indicated on the left hand side of the ms -attached below).

The part that i need clarity is on the Variance derivation i.e on the right hand side of the ms solution.

I know that Variance=npq, with n=1 and q=1-p this part is clear...but i do not seem to get the highlighted part.

 

[chwala]

I just saw the steps...one has to know the derivation though...it stems from

$Var (X)= E[X^2]-(E(X))^2$.

The derivation of Variance is as shown on the attachment below;

From here the steps to solution is straightforward.

$Var(X)=\dfrac{1-p}{p^2}$
now in our problem we shall therefore have;

$\dfrac{1}{p}=2×\dfrac{1-p}{p^2}$

$p=2-2p$

$3p=2$

$p=\dfrac{2}{3}$

Something new i am learning here today...i can see that it all refers to the geometric distribution...for the next part of the question, it asks us to find;

Here we shall use,

$Pr(X=i)=(1-p)^{i-1}p$

given that $p=\frac{2}{3}$

$Pr(X=4)=\left[\frac{1}{3}\right]^{3}×\frac{2}{3}=\frac{2}{81}$

Any other alternative approach or insight is highly welcome guys.

 

[pasmith]

I would write $$p \sum_{k=1}^\infty k^2 (1-p)^{k-1} = \frac{p}{1-p} F(1-p)$$ where $$F(x) = \sum_{k=0}^\infty k^2 x^k.$$ Since multiplying by $k$ inside the power series can be replaced by operating with $x\frac{d}{dx}$ outside it we get $$\begin{split} F(x) = \left(x \frac{d}{dx}\right)^2\sum_{k=0}^\infty x^k &= \left(x \frac{d}{dx}\right)^2\frac{1}{1-x} \\ &= \frac{x(x+1)}{(1 - x)^3}.\end{split}$$ Thus $$\frac{p}{1-p}F(1-p) = \frac{p}{1-p} \frac{(1-p)(2-p)}{p^3} = \frac{2-p}{p^2}$$ as required.

 

[PeroK]

Another method is to note that:
$$E(X^2) = p + (1-p)E((X+1)^2)$$$$ = p + (1-p)(E(X^2) +2E(X) + 1)$$$$ = (1-p)E(X^2) +\frac{2-p}{p}$$Hence:
$$E(X^2) = \frac {2-p}{p^2}$$Note that we can justify the "trick" in the first line using power series:
$$E(X^2) = \sum_{k=1}^{\infty}k^2p(1-p)^{k-1}$$$$ = p + \sum_{k=2}^{\infty}k^2p(1-p)^{k-1}$$$$=p+ (1-p)\sum_{k=2}^{\infty}k^2p(1-p)^{k-2}$$$$ = p+ (1-p)\sum_{k=1}^{\infty}(k+1)^2p(1-p)^{k-1}$$$$= p + (1-p)E((X+1)^2)$$