Answer to Question #348053 in Statistics and Probability for ria

Question #348053

Show that

f(x)=(1/2)^(x+1) , x = 0, 1, 2, 3, 4, 5,...

is a valid PMF for a discrete random variable. Also find

out its CDF.


1
Expert's answer
2022-06-06T16:51:44-0400

a.


x=0(12)x+1=1/211/2=1,True\displaystyle\sum_{x=0}^{\infin}(\dfrac{1}{2})^{x+1}=\dfrac{1/2}{1-1/2}=1, True

Therefore f(x)(12)x+1,x=0,1,2,3,4,5,...f(x)(\dfrac{1}{2})^{x+1}, x=0,1,2,3,4,5,... is a valid PMF for a discrete random variable.


b.


F(x)=yxxf(y)=y=0x(12)y+1=12y=0x(12)yF(x)=\displaystyle\sum_{y\le x}^{x}f(y)=\displaystyle\sum_{y=0}^{x}(\dfrac{1}{2})^{y+1}=\dfrac{1}{2}\displaystyle\sum_{y=0}^{x}(\dfrac{1}{2})^{y}

=12(1(1/2)x+111/2)=1(12)x+1=\dfrac{1}{2}(\dfrac{1-(1/2)^{x+1}}{1-1/2})=1-(\dfrac{1}{2})^{x+1}

F(x)={0x<01(12)x+1x=0,1,2,3,4,5,...F(x)= \begin{cases} 0&x<0 \\ 1-(\dfrac{1}{2})^{x+1}&x=0,1,2,3,4,5,... \end{cases}


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