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F(x) = (128/127)(1/2)x, x = 1,2,3,...7. determine the requested values: round your answers to three decimal places (e.g. 98.765). (a)p(x ≤ 1) (b)p(x > 1) (c) mean (d) variance

User TylerP
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a.

\mathbb P(X\le 1)=\mathbb P(X=1)=(128)/(127)\left(\frac12\right)^1=(64)/(127)

b.

\mathbb P(X>1)=1-\mathbb P(X\le1)=1-(64)/(127)=(63)/(127)

c.

\mathbb E(X)=\displaystyle\sum_(x=1)^7 x\,f_X(x)=(64)/(127)\sum_(x=1)^7 x\left(\frac12\right)^(x-1)

Suppose
f(y)=\displaystyle\sum_(x=0)^7 y^x. Then
f'(y)=\displaystyle\sum_(x=1)^7 xy^(x-1). So if we can find a closed form for
f(y), in terms of
y, we can find
\mathbb E(X) by evaluating the derivative of
f(y) at
y=\frac12.


f(y)=\displaystyle\sum_(x=0)^7 y^x=y^0+y^1+y^2+\cdots+y^6+y^7

y\,f(y)=y^1+y^2+y^3+\cdots+y^7+y^8

f(y)-y\,f(y)=y^0-y^8

(1-y)f(y)=1-y^8

f(y)=(1-y^8)/(1-y)

\implies f'(y)=(7y^8-8y^7+1)/((1-y)^2)

\implies\mathbb E(X)=(64)/(127)f'\left(\frac12\right)=(64)/(127)*(247)/(64)=(247)/(127)

d.

\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2

We find
\mathbb E(X^2) in a similar manner as in (c).


\mathbb E(X^2)=\displaystyle\sum_(x=1)^7 x^2\,f_X(x)=(32)/(127)\sum_(x=1)^7x^2\left(\frac12\right)^(x-2)

Now,


f(y)=\displaystyle\sum_(x=0)^7y^x

\implies f'(y)=\displaystyle\sum_(x=1)^7xy^(x-1)

\implies f''(y)=\displaystyle\sum_(x=2)^7x(x-1)y^(x-2)

We know that


f''(y)=-(42y^8-96y^7+56y^6-2)/((1-y)^3)

\implies f''\left(\frac12\right)=(219)/(16)

We also have


f''(y)=\displaystyle\sum_(x=2)^7x(x-1)y^(x-2)

f''(y)=\displaystyle\sum_(x=2)^7x^2y^(x-2)-\sum_(x=2)^7xy^(x-2)

f''(y)=\displaystyle\frac1{y^2}\left(\sum_(x=2)^7x^2y^x-\sum_(x=2)^7xy^x\right)

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)

f''(y)=\displaystyle\frac1{y^2}\left(\sum_(x=1)^7x^2y^x-\sum_(x=1)^7xy^x\right)

so that when
y=\frac12, we get


(219)/(16)=4\left((127)/(128)\mathbb E(X^2)-(127)/(128)\mathbb E(X)\right)\implies\mathbb E(X^2)=(685)/(127)

Then


\mathbb V(X)=(685)/(127)-\left((247)/(127)\right)^2=(25,986)/(16,129)
User Bondifrench
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