Solve the problem involving toss of a biased coin- Probability

  • Thread starter Thread starter chwala
  • Start date Start date
  • Tags Tags
    Probability
chwala
Gold Member
Messages
2,825
Reaction score
413
Homework Statement
See attached
Relevant Equations
Probability -Expectation and variance
This is the problem;
1662161507222.png


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.

1662162051838.png
 
Last edited:
Physics news on Phys.org
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;

1662208856314.png


1662208891514.png


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;

1662209756435.png

1662209983794.png
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.
 
Last edited:
I would write <br /> p \sum_{k=1}^\infty k^2 (1-p)^{k-1} = \frac{p}{1-p} F(1-p) where <br /> 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 <br /> \begin{split}<br /> F(x) = \left(x \frac{d}{dx}\right)^2\sum_{k=0}^\infty x^k &amp;= \left(x \frac{d}{dx}\right)^2\frac{1}{1-x} \\<br /> &amp;= \frac{x(x+1)}{(1 - x)^3}.\end{split} Thus <br /> \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.
 
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:
$$E(X^2) = \sum_{k=1}^{\infty}k^2p(1-p)^{k-1}$$$$ = p + \sum_{k=2}^{\infty}k^2p(1-p)^{k-1}$$$$=p+ (1-p)\sum_{k=2}^{\infty}k^2p(1-p)^{k-2}$$$$ = p+ (1-p)\sum_{k=1}^{\infty}(k+1)^2p(1-p)^{k-1}$$$$= p + (1-p)E((X+1)^2)$$
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top