How do we calculate the expected value of $X^{-1}$, where $X$ has geometric distribution?












0












$begingroup$


Let $X$ have a geometric distribution such that $p_{X}(x) = textbf{P}(X = x) = q^{x−1}p$, where $x geq 1$. Show that $textbf{E}(X^{-1}) =
log(p^{1/p−1})$.



MY ATTEMPT



According to the properties of the expectation, we have
begin{align*}
textbf{E}(g(X)) = sum_{x=1}^{infty}g(x)p_{X}(x)
end{align*}



In the particular case when $g(X) = X^{-1}$, the problem reduces to determine
begin{align*}
textbf{E}(X^{-1}) = sum_{x=1}^{infty}frac{q^{x-1}p}{x} = psum_{x=1}^{infty}frac{q^{x-1}}{x}
end{align*}



Now consider the series



begin{align*}
f(w) & = 1 + frac{w}{2} + frac{w^{2}}{3} + ldots Rightarrow F(w) = int_{0}^{w}f(u)mathrm{d}u = frac{w}{1^{2}} + frac{w^{2}}{2^{2}} + frac{w^{3}}{3^{2}} + ldots
end{align*}



If I am on the right track, $textbf{E}(X^{-1}) = pF^{prime}(1-p)$. However, I am unable to find out the closed formula for both $f$ and $F$. Am I failing at some point? Otherwise, could someone help me out? Thanks!



EDIT



According to Alex's contribution, we can rewrite the given expression as
begin{align*}
textbf{E}(X^{-1}) = frac{p}{q}sum_{x=1}^{infty}frac{q^{x}}{x}
end{align*}



Therefore, if we conveniently define
begin{align*}
G(w) & = w + frac{w^{2}}{2} + frac{w^{3}}{3} + ldots Rightarrow G^{prime}(w) = g(w) = 1 + w + w^{2} + ldots = frac{1}{1-w} Rightarrow\\
G(w) & = log(1-w) Rightarrow textbf{E}(X^{-1}) = frac{p}{q}G(q) = frac{p}{1-p}log(1 - (1 - p)) = frac{p}{1-p}log(p)
end{align*}



Am I missing something?






probability proof-verification probability-distributions expected-value







share|cite|improve this question











$endgroup$












  • $begingroup$
    You have the right sort of idea. Try using $psum_{x = 1}^{infty} frac{q^{x - 1}}{x} = frac{p}{q} sum_{x = 1}^{infty} int_{0}^{q}u^{x - 1} du = frac{p}{q}int_{0}^{q} sum_{x = 1}^{infty} u^{x - 1} du$
    $endgroup$
    – Alex
    Jan 25 at 23:11












  • $begingroup$
    You're missing a minus sign. $E(X^{-1})=-log(p^{1/p-1})$.
    $endgroup$
    – JimB
    Jan 26 at 1:56












  • $begingroup$
    Thanks for the contribution!
    $endgroup$
    – user1337
    Jan 26 at 1:58


















0












$begingroup$


Let $X$ have a geometric distribution such that $p_{X}(x) = textbf{P}(X = x) = q^{x−1}p$, where $x geq 1$. Show that $textbf{E}(X^{-1}) =
log(p^{1/p−1})$.



MY ATTEMPT



According to the properties of the expectation, we have
begin{align*}
textbf{E}(g(X)) = sum_{x=1}^{infty}g(x)p_{X}(x)
end{align*}



In the particular case when $g(X) = X^{-1}$, the problem reduces to determine
begin{align*}
textbf{E}(X^{-1}) = sum_{x=1}^{infty}frac{q^{x-1}p}{x} = psum_{x=1}^{infty}frac{q^{x-1}}{x}
end{align*}



Now consider the series



begin{align*}
f(w) & = 1 + frac{w}{2} + frac{w^{2}}{3} + ldots Rightarrow F(w) = int_{0}^{w}f(u)mathrm{d}u = frac{w}{1^{2}} + frac{w^{2}}{2^{2}} + frac{w^{3}}{3^{2}} + ldots
end{align*}



If I am on the right track, $textbf{E}(X^{-1}) = pF^{prime}(1-p)$. However, I am unable to find out the closed formula for both $f$ and $F$. Am I failing at some point? Otherwise, could someone help me out? Thanks!



EDIT



According to Alex's contribution, we can rewrite the given expression as
begin{align*}
textbf{E}(X^{-1}) = frac{p}{q}sum_{x=1}^{infty}frac{q^{x}}{x}
end{align*}



Therefore, if we conveniently define
begin{align*}
G(w) & = w + frac{w^{2}}{2} + frac{w^{3}}{3} + ldots Rightarrow G^{prime}(w) = g(w) = 1 + w + w^{2} + ldots = frac{1}{1-w} Rightarrow\\
G(w) & = log(1-w) Rightarrow textbf{E}(X^{-1}) = frac{p}{q}G(q) = frac{p}{1-p}log(1 - (1 - p)) = frac{p}{1-p}log(p)
end{align*}



Am I missing something?






probability proof-verification probability-distributions expected-value







share|cite|improve this question











$endgroup$












  • $begingroup$
    You have the right sort of idea. Try using $psum_{x = 1}^{infty} frac{q^{x - 1}}{x} = frac{p}{q} sum_{x = 1}^{infty} int_{0}^{q}u^{x - 1} du = frac{p}{q}int_{0}^{q} sum_{x = 1}^{infty} u^{x - 1} du$
    $endgroup$
    – Alex
    Jan 25 at 23:11












  • $begingroup$
    You're missing a minus sign. $E(X^{-1})=-log(p^{1/p-1})$.
    $endgroup$
    – JimB
    Jan 26 at 1:56












  • $begingroup$
    Thanks for the contribution!
    $endgroup$
    – user1337
    Jan 26 at 1:58
















0












0








0





$begingroup$


Let $X$ have a geometric distribution such that $p_{X}(x) = textbf{P}(X = x) = q^{x−1}p$, where $x geq 1$. Show that $textbf{E}(X^{-1}) =
log(p^{1/p−1})$.



MY ATTEMPT



According to the properties of the expectation, we have
begin{align*}
textbf{E}(g(X)) = sum_{x=1}^{infty}g(x)p_{X}(x)
end{align*}



In the particular case when $g(X) = X^{-1}$, the problem reduces to determine
begin{align*}
textbf{E}(X^{-1}) = sum_{x=1}^{infty}frac{q^{x-1}p}{x} = psum_{x=1}^{infty}frac{q^{x-1}}{x}
end{align*}



Now consider the series



begin{align*}
f(w) & = 1 + frac{w}{2} + frac{w^{2}}{3} + ldots Rightarrow F(w) = int_{0}^{w}f(u)mathrm{d}u = frac{w}{1^{2}} + frac{w^{2}}{2^{2}} + frac{w^{3}}{3^{2}} + ldots
end{align*}



If I am on the right track, $textbf{E}(X^{-1}) = pF^{prime}(1-p)$. However, I am unable to find out the closed formula for both $f$ and $F$. Am I failing at some point? Otherwise, could someone help me out? Thanks!



EDIT



According to Alex's contribution, we can rewrite the given expression as
begin{align*}
textbf{E}(X^{-1}) = frac{p}{q}sum_{x=1}^{infty}frac{q^{x}}{x}
end{align*}



Therefore, if we conveniently define
begin{align*}
G(w) & = w + frac{w^{2}}{2} + frac{w^{3}}{3} + ldots Rightarrow G^{prime}(w) = g(w) = 1 + w + w^{2} + ldots = frac{1}{1-w} Rightarrow\\
G(w) & = log(1-w) Rightarrow textbf{E}(X^{-1}) = frac{p}{q}G(q) = frac{p}{1-p}log(1 - (1 - p)) = frac{p}{1-p}log(p)
end{align*}



Am I missing something?






probability proof-verification probability-distributions expected-value







share|cite|improve this question











$endgroup$




Let $X$ have a geometric distribution such that $p_{X}(x) = textbf{P}(X = x) = q^{x−1}p$, where $x geq 1$. Show that $textbf{E}(X^{-1}) =
log(p^{1/p−1})$.



MY ATTEMPT



According to the properties of the expectation, we have
begin{align*}
textbf{E}(g(X)) = sum_{x=1}^{infty}g(x)p_{X}(x)
end{align*}



In the particular case when $g(X) = X^{-1}$, the problem reduces to determine
begin{align*}
textbf{E}(X^{-1}) = sum_{x=1}^{infty}frac{q^{x-1}p}{x} = psum_{x=1}^{infty}frac{q^{x-1}}{x}
end{align*}



Now consider the series



begin{align*}
f(w) & = 1 + frac{w}{2} + frac{w^{2}}{3} + ldots Rightarrow F(w) = int_{0}^{w}f(u)mathrm{d}u = frac{w}{1^{2}} + frac{w^{2}}{2^{2}} + frac{w^{3}}{3^{2}} + ldots
end{align*}



If I am on the right track, $textbf{E}(X^{-1}) = pF^{prime}(1-p)$. However, I am unable to find out the closed formula for both $f$ and $F$. Am I failing at some point? Otherwise, could someone help me out? Thanks!



EDIT



According to Alex's contribution, we can rewrite the given expression as
begin{align*}
textbf{E}(X^{-1}) = frac{p}{q}sum_{x=1}^{infty}frac{q^{x}}{x}
end{align*}



Therefore, if we conveniently define
begin{align*}
G(w) & = w + frac{w^{2}}{2} + frac{w^{3}}{3} + ldots Rightarrow G^{prime}(w) = g(w) = 1 + w + w^{2} + ldots = frac{1}{1-w} Rightarrow\\
G(w) & = log(1-w) Rightarrow textbf{E}(X^{-1}) = frac{p}{q}G(q) = frac{p}{1-p}log(1 - (1 - p)) = frac{p}{1-p}log(p)
end{align*}



Am I missing something?






probability proof-verification probability-distributions expected-value




probability proof-verification probability-distributions expected-value






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 26 at 0:54







user1337

















asked Jan 25 at 22:58









user1337user1337

46110




46110












  • $begingroup$
    You have the right sort of idea. Try using $psum_{x = 1}^{infty} frac{q^{x - 1}}{x} = frac{p}{q} sum_{x = 1}^{infty} int_{0}^{q}u^{x - 1} du = frac{p}{q}int_{0}^{q} sum_{x = 1}^{infty} u^{x - 1} du$
    $endgroup$
    – Alex
    Jan 25 at 23:11












  • $begingroup$
    You're missing a minus sign. $E(X^{-1})=-log(p^{1/p-1})$.
    $endgroup$
    – JimB
    Jan 26 at 1:56












  • $begingroup$
    Thanks for the contribution!
    $endgroup$
    – user1337
    Jan 26 at 1:58




















  • $begingroup$
    You have the right sort of idea. Try using $psum_{x = 1}^{infty} frac{q^{x - 1}}{x} = frac{p}{q} sum_{x = 1}^{infty} int_{0}^{q}u^{x - 1} du = frac{p}{q}int_{0}^{q} sum_{x = 1}^{infty} u^{x - 1} du$
    $endgroup$
    – Alex
    Jan 25 at 23:11












  • $begingroup$
    You're missing a minus sign. $E(X^{-1})=-log(p^{1/p-1})$.
    $endgroup$
    – JimB
    Jan 26 at 1:56












  • $begingroup$
    Thanks for the contribution!
    $endgroup$
    – user1337
    Jan 26 at 1:58


















$begingroup$
You have the right sort of idea. Try using $psum_{x = 1}^{infty} frac{q^{x - 1}}{x} = frac{p}{q} sum_{x = 1}^{infty} int_{0}^{q}u^{x - 1} du = frac{p}{q}int_{0}^{q} sum_{x = 1}^{infty} u^{x - 1} du$
$endgroup$
– Alex
Jan 25 at 23:11






$begingroup$
You have the right sort of idea. Try using $psum_{x = 1}^{infty} frac{q^{x - 1}}{x} = frac{p}{q} sum_{x = 1}^{infty} int_{0}^{q}u^{x - 1} du = frac{p}{q}int_{0}^{q} sum_{x = 1}^{infty} u^{x - 1} du$
$endgroup$
– Alex
Jan 25 at 23:11














$begingroup$
You're missing a minus sign. $E(X^{-1})=-log(p^{1/p-1})$.
$endgroup$
– JimB
Jan 26 at 1:56






$begingroup$
You're missing a minus sign. $E(X^{-1})=-log(p^{1/p-1})$.
$endgroup$
– JimB
Jan 26 at 1:56














$begingroup$
Thanks for the contribution!
$endgroup$
– user1337
Jan 26 at 1:58






$begingroup$
Thanks for the contribution!
$endgroup$
– user1337
Jan 26 at 1:58












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