a.
[tex]\mathbb P(X\le 1)=\mathbb P(X=1)=\dfrac{128}{127}\left(\dfrac12\right)^1=\dfrac{64}{127}[/tex]
b.
[tex]\mathbb P(X>1)=1-\mathbb P(X\le1)=1-\dfrac{64}{127}=\dfrac{63}{127}[/tex]
c.
[tex]\mathbb E(X)=\displaystyle\sum_{x=1}^7 x\,f_X(x)=\frac{64}{127}\sum_{x=1}^7 x\left(\frac12\right)^{x-1}[/tex]
Suppose [tex]f(y)=\displaystyle\sum_{x=0}^7 y^x[/tex]. Then [tex]f'(y)=\displaystyle\sum_{x=1}^7 xy^{x-1}[/tex]. So if we can find a closed form for [tex]f(y)[/tex], in terms of [tex]y[/tex], we can find [tex]\mathbb E(X)[/tex] by evaluating the derivative of [tex]f(y)[/tex] at [tex]y=\dfrac12[/tex].
[tex]f(y)=\displaystyle\sum_{x=0}^7 y^x=y^0+y^1+y^2+\cdots+y^6+y^7[/tex]
[tex]y\,f(y)=y^1+y^2+y^3+\cdots+y^7+y^8[/tex]
[tex]f(y)-y\,f(y)=y^0-y^8[/tex]
[tex](1-y)f(y)=1-y^8[/tex]
[tex]f(y)=\dfrac{1-y^8}{1-y}[/tex]
[tex]\implies f'(y)=\dfrac{7y^8-8y^7+1}{(1-y)^2}[/tex]
[tex]\implies\mathbb E(X)=\dfrac{64}{127}f'\left(\dfrac12\right)=\dfrac{64}{127}\times\dfrac{247}{64}=\dfrac{247}{127}[/tex]
d.
[tex]\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2[/tex]
We find [tex]\mathbb E(X^2)[/tex] in a similar manner as in (c).
[tex]\mathbb E(X^2)=\displaystyle\sum_{x=1}^7 x^2\,f_X(x)=\frac{32}{127}\sum_{x=1}^7x^2\left(\frac12\right)^{x-2}[/tex]
Now,
[tex]f(y)=\displaystyle\sum_{x=0}^7y^x[/tex]
[tex]\implies f'(y)=\displaystyle\sum_{x=1}^7xy^{x-1}[/tex]
[tex]\implies f''(y)=\displaystyle\sum_{x=2}^7x(x-1)y^{x-2}[/tex]
We know that
[tex]f''(y)=-\dfrac{42y^8-96y^7+56y^6-2}{(1-y)^3}[/tex]
[tex]\implies f''\left(\dfrac12\right)=\dfrac{219}{16}[/tex]
We also have
[tex]f''(y)=\displaystyle\sum_{x=2}^7x(x-1)y^{x-2}[/tex]
[tex]f''(y)=\displaystyle\sum_{x=2}^7x^2y^{x-2}-\sum_{x=2}^7xy^{x-2}[/tex]
[tex]f''(y)=\displaystyle\frac1{y^2}\left(\sum_{x=2}^7x^2y^x-\sum_{x=2}^7xy^x\right)[/tex]
[tex]f''(y)=\displaystyle\frac1{y^2}\left(\bigg(\sum_{x=1}^7x^2y^x-y\bigg)-\bigg(\sum_{x=1}^7xy^x-y\bigg)\right)[/tex]
[tex]f''(y)=\displaystyle\frac1{y^2}\left(\sum_{x=1}^7x^2y^x-\sum_{x=1}^7xy^x\right)[/tex]
so that when [tex]y=\dfrac12[/tex], we get
[tex]\dfrac{219}{16}=4\left(\dfrac{127}{128}\mathbb E(X^2)-\dfrac{127}{128}\mathbb E(X)\right)\implies\mathbb E(X^2)=\dfrac{685}{127}[/tex]
Then
[tex]\mathbb V(X)=\dfrac{685}{127}-\left(\dfrac{247}{127}\right)^2=\dfrac{25,986}{16,129}[/tex]
