The probability mass function of for W is given by
, the expected value(e(w)) is
and the expressions for
is given by ![\frac{1}{2}^{2k-2},K=1,2,3...](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5E%7B2k-2%7D%2CK%3D1%2C2%2C3...)
To solve this problem we have to understand pmf or probability mass function and e(W) or expected value.
Probability mass function is defined as the function which given the probability of a discreet random variable.
The weighted average of the possible values of a random variable with weights given by their respective probabilities is known as expected value.
Jon flips a fair coin until a head is witnessed, and independently Anna Flips a fair coin(a different coin than Jon) until a head is witnessed. Let W be the number of flips between Jon and Anna.
Let us now define the random variable
X: Number of flips done by Jon
Y: Number of flips done by Mocus
So, PMF of X
![P(X=x)=\frac{1}{2}\times(1-\frac{1}{2})^{(x-1)},x=1,2,...\\P(X=x)=(\frac{1}{2})^x\\](https://tex.z-dn.net/?f=P%28X%3Dx%29%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%281-%5Cfrac%7B1%7D%7B2%7D%29%5E%7B%28x-1%29%7D%2Cx%3D1%2C2%2C...%5C%5CP%28X%3Dx%29%3D%28%5Cfrac%7B1%7D%7B2%7D%29%5Ex%5C%5C)
![P(X\geq x)=\sum_{i=x}^{\infty}(\frac{1}{2})^i\\P(X\geq x)=(\frac{1}{2})^x+(\frac{1}{2})^{x+1}+.....\\P(X\geq x)=\frac{(\frac{1}{2})^x}{1-\frac{1}{2}}\\P(X\geq x)=(\frac{1}{2})^{x-1}\\](https://tex.z-dn.net/?f=P%28X%5Cgeq%20x%29%3D%5Csum_%7Bi%3Dx%7D%5E%7B%5Cinfty%7D%28%5Cfrac%7B1%7D%7B2%7D%29%5Ei%5C%5CP%28X%5Cgeq%20x%29%3D%28%5Cfrac%7B1%7D%7B2%7D%29%5Ex%2B%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7Bx%2B1%7D%2B.....%5C%5CP%28X%5Cgeq%20x%29%3D%5Cfrac%7B%28%5Cfrac%7B1%7D%7B2%7D%29%5Ex%7D%7B1-%5Cfrac%7B1%7D%7B2%7D%7D%5C%5CP%28X%5Cgeq%20x%29%3D%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7Bx-1%7D%5C%5C)
Simmilarly
![P(Y\geq y)=(\frac{1}{2})^{(y-1)},y=1,2,3....\\](https://tex.z-dn.net/?f=P%28Y%5Cgeq%20y%29%3D%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7B%28y-1%29%7D%2Cy%3D1%2C2%2C3....%5C%5C)
Now W=min(X,Y)
So,
![P(w\geq K)\\=P(min(X,Y)\geq K)\\=P(X\geq K,Y\geq K)\\=(\frac{1}{2})^{K-1}(\frac{1}{2})^{K-1}\\=(\frac{1}{2})^{2K-2}](https://tex.z-dn.net/?f=P%28w%5Cgeq%20K%29%5C%5C%3DP%28min%28X%2CY%29%5Cgeq%20K%29%5C%5C%3DP%28X%5Cgeq%20K%2CY%5Cgeq%20K%29%5C%5C%3D%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7BK-1%7D%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7BK-1%7D%5C%5C%3D%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7B2K-2%7D)
a) Now let us derive the probability mass function(pmf) for W:
![P(W=K)=P(W\geq K)-P(W\geq K+1)\\P(W=K)=\frac{3}{2^{2K}}](https://tex.z-dn.net/?f=P%28W%3DK%29%3DP%28W%5Cgeq%20K%29-P%28W%5Cgeq%20K%2B1%29%5C%5CP%28W%3DK%29%3D%5Cfrac%7B3%7D%7B2%5E%7B2K%7D%7D)
b) Let us calculate the expected value or e(W)
![E(W)=\sum^{\infty}_{i=1}ip(W=0)](https://tex.z-dn.net/?f=E%28W%29%3D%5Csum%5E%7B%5Cinfty%7D_%7Bi%3D1%7Dip%28W%3D0%29)
Solving we get
![=\sum^{\infty}_{I=1}P(W\geq i)\\=\sum^{\infty}_{I=1}\frac{1}{2^{2i-2}}\\=1+\frac{1}{4}+\frac{1}{16}+....\\=\frac{4}{3}](https://tex.z-dn.net/?f=%3D%5Csum%5E%7B%5Cinfty%7D_%7BI%3D1%7DP%28W%5Cgeq%20i%29%5C%5C%3D%5Csum%5E%7B%5Cinfty%7D_%7BI%3D1%7D%5Cfrac%7B1%7D%7B2%5E%7B2i-2%7D%7D%5C%5C%3D1%2B%5Cfrac%7B1%7D%7B4%7D%2B%5Cfrac%7B1%7D%7B16%7D%2B....%5C%5C%3D%5Cfrac%7B4%7D%7B3%7D)
c) The expression for
for K=1,2,3,4....
![P(W\geq K)\\=\frac{1}{2}^{2k-2},K=1,2,3...](https://tex.z-dn.net/?f=P%28W%5Cgeq%20K%29%5C%5C%3D%5Cfrac%7B1%7D%7B2%7D%5E%7B2k-2%7D%2CK%3D1%2C2%2C3...)
To learn more about probability mass function:
brainly.com/question/14263946
To learn more about expected value:
brainly.com/question/28197299
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