Mixing
Alternative approaches to obtain the expected value of the geometric distribution
$begingroup$
Given that $X$ has geometric distribution with $p_{X}(x) = p(1-p)^{x-1}$, determine $textbf{E}(X)$.
MY ATTEMPT
begin{align*}
textbf{E}(X) = sum_{x=1}^{infty}xp(1-p)^{x-1} = psum_{x=1}^{infty}x(1-p)^{x-1}
end{align*}
If we denote by
begin{align*}
F(w) = sum_{k=1}^{infty} w^{k} = frac{w}{1 - w}quadtext{for}quad |w| < 1
end{align*}
We conclude that
begin{align*}
textbf{E}(X) = psum_{x=1}^{infty}x(1-p)^{x-1} = pF^{prime}(1-p)
end{align*}
Since $displaystyle F^{prime}(w) = frac{1}{(1-w)^{2}}$, it is now possible to obtain the desired result
begin{align*}
textbf{E}(X) = frac{p}{(1-(1-p))^{2}} = frac{1}{p}
end{align*}
In the case that my answer is correct, could someone provide me any other approach to this problem? I'd prefer solutions which do not involve sophisticated methods. Thanks in advance.
probability probability-theory proof-verification probability-distributions
asked Jan 22 at 21:10
user1337user1337
46110
$endgroup$
add a comment |
$begingroup$
Given that $X$ has geometric distribution with $p_{X}(x) = p(1-p)^{x-1}$, determine $textbf{E}(X)$.
MY ATTEMPT
begin{align*}
textbf{E}(X) = sum_{x=1}^{infty}xp(1-p)^{x-1} = psum_{x=1}^{infty}x(1-p)^{x-1}
end{align*}
If we denote by
begin{align*}
F(w) = sum_{k=1}^{infty} w^{k} = frac{w}{1 - w}quadtext{for}quad |w| < 1
end{align*}
We conclude that
begin{align*}
textbf{E}(X) = psum_{x=1}^{infty}x(1-p)^{x-1} = pF^{prime}(1-p)
end{align*}
Since $displaystyle F^{prime}(w) = frac{1}{(1-w)^{2}}$, it is now possible to obtain the desired result
begin{align*}
textbf{E}(X) = frac{p}{(1-(1-p))^{2}} = frac{1}{p}
end{align*}
In the case that my answer is correct, could someone provide me any other approach to this problem? I'd prefer solutions which do not involve sophisticated methods. Thanks in advance.
probability probability-theory proof-verification probability-distributions
asked Jan 22 at 21:10
user1337user1337
46110
$endgroup$
add a comment |
$begingroup$
Given that $X$ has geometric distribution with $p_{X}(x) = p(1-p)^{x-1}$, determine $textbf{E}(X)$.
MY ATTEMPT
begin{align*}
textbf{E}(X) = sum_{x=1}^{infty}xp(1-p)^{x-1} = psum_{x=1}^{infty}x(1-p)^{x-1}
end{align*}
If we denote by
begin{align*}
F(w) = sum_{k=1}^{infty} w^{k} = frac{w}{1 - w}quadtext{for}quad |w| < 1
end{align*}
We conclude that
begin{align*}
textbf{E}(X) = psum_{x=1}^{infty}x(1-p)^{x-1} = pF^{prime}(1-p)
end{align*}
Since $displaystyle F^{prime}(w) = frac{1}{(1-w)^{2}}$, it is now possible to obtain the desired result
begin{align*}
textbf{E}(X) = frac{p}{(1-(1-p))^{2}} = frac{1}{p}
end{align*}
In the case that my answer is correct, could someone provide me any other approach to this problem? I'd prefer solutions which do not involve sophisticated methods. Thanks in advance.
probability probability-theory proof-verification probability-distributions
asked Jan 22 at 21:10
user1337user1337
46110
$endgroup$
Given that $X$ has geometric distribution with $p_{X}(x) = p(1-p)^{x-1}$, determine $textbf{E}(X)$.
MY ATTEMPT
begin{align*}
textbf{E}(X) = sum_{x=1}^{infty}xp(1-p)^{x-1} = psum_{x=1}^{infty}x(1-p)^{x-1}
end{align*}
If we denote by
begin{align*}
F(w) = sum_{k=1}^{infty} w^{k} = frac{w}{1 - w}quadtext{for}quad |w| < 1
end{align*}
We conclude that
begin{align*}
textbf{E}(X) = psum_{x=1}^{infty}x(1-p)^{x-1} = pF^{prime}(1-p)
end{align*}
Since $displaystyle F^{prime}(w) = frac{1}{(1-w)^{2}}$, it is now possible to obtain the desired result
begin{align*}
textbf{E}(X) = frac{p}{(1-(1-p))^{2}} = frac{1}{p}
end{align*}
In the case that my answer is correct, could someone provide me any other approach to this problem? I'd prefer solutions which do not involve sophisticated methods. Thanks in advance.
probability probability-theory proof-verification probability-distributions
probability probability-theory proof-verification probability-distributions
asked Jan 22 at 21:10
user1337user1337
46110
asked Jan 22 at 21:10
user1337user1337
46110
edited Jan 22 at 22:47
user1337
asked Jan 22 at 21:10
user1337user1337
46110
asked Jan 22 at 21:10
user1337user1337
46110
asked Jan 22 at 21:10
user1337user1337
46110
46110
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The geometric distribution gives the number of trials until the first success (including the successful one, in your mass function above) in a sequence of trials with probability of success $p$.
The first trial is either a success (probability $p$) or a failure (probability $1-p$); if it is a success, you are done and $X=1$. If it is a failure, you are left with another geometric process which you must add one extra failure to.
In other words,
$$
mathbb{E}[X]=pcdot1+(1-p)cdot(1+mathbb{E}[X])=1+(1-p)mathbb{E}[X].
$$
Subtracting $(1-p)mathbb{E}[X]$ from both sides yields
$$
mathbb{E}[X]-(1-p)mathbb{E}[X]=1,
$$
which simplifies to
$$
pmathbb{E}[X]=1qquadRightarrowqquadmathbb{E}[X]=frac{1}{p}.
$$
answered Jan 22 at 22:56
Nick PetersonNick Peterson
26.8k23962
$endgroup$
$begingroup$
Nice argument! Thanks for the contribution.
$endgroup$
– user1337
Jan 22 at 23:04
add a comment |
$begingroup$
Your result is correct. You can also use a version of Fubini-Tonelli's theorem (i.e. changing order of summation):
$$begin{eqnarray}
sum_{j=0}^infty j(1-p)^{j-1}&=&sum_{j=0}^infty left(sum_{k=0}^{j-1} 1right)(1-p)^{j-1}\&=&sum_{k=0}^infty left(sum_{j=k+1}^infty(1-p)^{j-1} right)\
&=&sum_{k=0}^infty frac{(1-p)^k}{p}\
&=&frac{1}{p^2}.
end{eqnarray}$$ This gives
$$
Bbb E[X]=psum_{j=0}^infty j(1-p)^{j-1}=frac{1}{p}.
$$
Note: Note that $$binom{-2}{j}=frac{(-2)(-3)cdots(-1-j)}{j!}=(-1)^j(j+1).$$ Generalized binomial theorem also gives
$$
sum_{j=0}^infty (j+1)x^j=sum_{j=0}^infty binom{-2}{j}(-x)^j=(1-x)^{-2}.
$$
answered Jan 22 at 21:24
SongSong
17k21145
$endgroup$
add a comment |
$begingroup$
I can offer a slightly different way of doing this
$$ E(X) = p sum_{x=1}^{infty} x(1-p)^{x-1} $$
$$ (1-p)E(X) =psum_{x=1}^{infty} x(1-p)^{x} $$
$$ E(X) - (1-p)E(X) = psum_{x=0}^{infty} (1-p)^{x} $$
You can see this writing down the first few terms in each sum :
$$ E(X) = p( 1 + 2(1-p) + 3(1-p)^2 ...)$$
$$ (1-p)E(X) = p( (1-p)+ 2(1-p)^2 + 3(1-p)^3 ...) $$
$$ E(X) - (1-p)E(X) =p (1+ (1-p) + (1-p)^2 + (1-p)^3 ...) $$
$$ E(X) - (1-p)E(X) = p frac{1}{1-(1-p)} $$
$$ E(X) =frac{1}{p}$$
answered Jan 22 at 21:52
Rohan NuckchadyRohan Nuckchady
1111
$endgroup$
add a comment |
Your Answer
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3 Answers
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active
oldest
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3 Answers
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$begingroup$
The geometric distribution gives the number of trials until the first success (including the successful one, in your mass function above) in a sequence of trials with probability of success $p$.
The first trial is either a success (probability $p$) or a failure (probability $1-p$); if it is a success, you are done and $X=1$. If it is a failure, you are left with another geometric process which you must add one extra failure to.
In other words,
$$
mathbb{E}[X]=pcdot1+(1-p)cdot(1+mathbb{E}[X])=1+(1-p)mathbb{E}[X].
$$
Subtracting $(1-p)mathbb{E}[X]$ from both sides yields
$$
mathbb{E}[X]-(1-p)mathbb{E}[X]=1,
$$
which simplifies to
$$
pmathbb{E}[X]=1qquadRightarrowqquadmathbb{E}[X]=frac{1}{p}.
$$
answered Jan 22 at 22:56
Nick PetersonNick Peterson
26.8k23962
$endgroup$
$begingroup$
Nice argument! Thanks for the contribution.
$endgroup$
– user1337
Jan 22 at 23:04
add a comment |
$begingroup$
The geometric distribution gives the number of trials until the first success (including the successful one, in your mass function above) in a sequence of trials with probability of success $p$.
The first trial is either a success (probability $p$) or a failure (probability $1-p$); if it is a success, you are done and $X=1$. If it is a failure, you are left with another geometric process which you must add one extra failure to.
In other words,
$$
mathbb{E}[X]=pcdot1+(1-p)cdot(1+mathbb{E}[X])=1+(1-p)mathbb{E}[X].
$$
Subtracting $(1-p)mathbb{E}[X]$ from both sides yields
$$
mathbb{E}[X]-(1-p)mathbb{E}[X]=1,
$$
which simplifies to
$$
pmathbb{E}[X]=1qquadRightarrowqquadmathbb{E}[X]=frac{1}{p}.
$$
answered Jan 22 at 22:56
Nick PetersonNick Peterson
26.8k23962
$endgroup$
$begingroup$
Nice argument! Thanks for the contribution.
$endgroup$
– user1337
Jan 22 at 23:04
add a comment |
$begingroup$
The geometric distribution gives the number of trials until the first success (including the successful one, in your mass function above) in a sequence of trials with probability of success $p$.
The first trial is either a success (probability $p$) or a failure (probability $1-p$); if it is a success, you are done and $X=1$. If it is a failure, you are left with another geometric process which you must add one extra failure to.
In other words,
$$
mathbb{E}[X]=pcdot1+(1-p)cdot(1+mathbb{E}[X])=1+(1-p)mathbb{E}[X].
$$
Subtracting $(1-p)mathbb{E}[X]$ from both sides yields
$$
mathbb{E}[X]-(1-p)mathbb{E}[X]=1,
$$
which simplifies to
$$
pmathbb{E}[X]=1qquadRightarrowqquadmathbb{E}[X]=frac{1}{p}.
$$
answered Jan 22 at 22:56
Nick PetersonNick Peterson
26.8k23962
$endgroup$
The geometric distribution gives the number of trials until the first success (including the successful one, in your mass function above) in a sequence of trials with probability of success $p$.
The first trial is either a success (probability $p$) or a failure (probability $1-p$); if it is a success, you are done and $X=1$. If it is a failure, you are left with another geometric process which you must add one extra failure to.
In other words,
$$
mathbb{E}[X]=pcdot1+(1-p)cdot(1+mathbb{E}[X])=1+(1-p)mathbb{E}[X].
$$
Subtracting $(1-p)mathbb{E}[X]$ from both sides yields
$$
mathbb{E}[X]-(1-p)mathbb{E}[X]=1,
$$
which simplifies to
$$
pmathbb{E}[X]=1qquadRightarrowqquadmathbb{E}[X]=frac{1}{p}.
$$
answered Jan 22 at 22:56
Nick PetersonNick Peterson
26.8k23962
answered Jan 22 at 22:56
Nick PetersonNick Peterson
26.8k23962
answered Jan 22 at 22:56
Nick PetersonNick Peterson
26.8k23962
answered Jan 22 at 22:56
Nick PetersonNick Peterson
26.8k23962
26.8k23962
$begingroup$
Nice argument! Thanks for the contribution.
$endgroup$
– user1337
Jan 22 at 23:04
add a comment |
$begingroup$
Nice argument! Thanks for the contribution.
$endgroup$
– user1337
Jan 22 at 23:04
$begingroup$
Nice argument! Thanks for the contribution.
$endgroup$
– user1337
Jan 22 at 23:04
$begingroup$
Nice argument! Thanks for the contribution.
$endgroup$
– user1337
Jan 22 at 23:04
add a comment |
$begingroup$
Your result is correct. You can also use a version of Fubini-Tonelli's theorem (i.e. changing order of summation):
$$begin{eqnarray}
sum_{j=0}^infty j(1-p)^{j-1}&=&sum_{j=0}^infty left(sum_{k=0}^{j-1} 1right)(1-p)^{j-1}\&=&sum_{k=0}^infty left(sum_{j=k+1}^infty(1-p)^{j-1} right)\
&=&sum_{k=0}^infty frac{(1-p)^k}{p}\
&=&frac{1}{p^2}.
end{eqnarray}$$ This gives
$$
Bbb E[X]=psum_{j=0}^infty j(1-p)^{j-1}=frac{1}{p}.
$$
Note: Note that $$binom{-2}{j}=frac{(-2)(-3)cdots(-1-j)}{j!}=(-1)^j(j+1).$$ Generalized binomial theorem also gives
$$
sum_{j=0}^infty (j+1)x^j=sum_{j=0}^infty binom{-2}{j}(-x)^j=(1-x)^{-2}.
$$
answered Jan 22 at 21:24
SongSong
17k21145
$endgroup$
add a comment |
$begingroup$
Your result is correct. You can also use a version of Fubini-Tonelli's theorem (i.e. changing order of summation):
$$begin{eqnarray}
sum_{j=0}^infty j(1-p)^{j-1}&=&sum_{j=0}^infty left(sum_{k=0}^{j-1} 1right)(1-p)^{j-1}\&=&sum_{k=0}^infty left(sum_{j=k+1}^infty(1-p)^{j-1} right)\
&=&sum_{k=0}^infty frac{(1-p)^k}{p}\
&=&frac{1}{p^2}.
end{eqnarray}$$ This gives
$$
Bbb E[X]=psum_{j=0}^infty j(1-p)^{j-1}=frac{1}{p}.
$$
Note: Note that $$binom{-2}{j}=frac{(-2)(-3)cdots(-1-j)}{j!}=(-1)^j(j+1).$$ Generalized binomial theorem also gives
$$
sum_{j=0}^infty (j+1)x^j=sum_{j=0}^infty binom{-2}{j}(-x)^j=(1-x)^{-2}.
$$
answered Jan 22 at 21:24
SongSong
17k21145
$endgroup$
add a comment |
$begingroup$
Your result is correct. You can also use a version of Fubini-Tonelli's theorem (i.e. changing order of summation):
$$begin{eqnarray}
sum_{j=0}^infty j(1-p)^{j-1}&=&sum_{j=0}^infty left(sum_{k=0}^{j-1} 1right)(1-p)^{j-1}\&=&sum_{k=0}^infty left(sum_{j=k+1}^infty(1-p)^{j-1} right)\
&=&sum_{k=0}^infty frac{(1-p)^k}{p}\
&=&frac{1}{p^2}.
end{eqnarray}$$ This gives
$$
Bbb E[X]=psum_{j=0}^infty j(1-p)^{j-1}=frac{1}{p}.
$$
Note: Note that $$binom{-2}{j}=frac{(-2)(-3)cdots(-1-j)}{j!}=(-1)^j(j+1).$$ Generalized binomial theorem also gives
$$
sum_{j=0}^infty (j+1)x^j=sum_{j=0}^infty binom{-2}{j}(-x)^j=(1-x)^{-2}.
$$
answered Jan 22 at 21:24
SongSong
17k21145
$endgroup$
Your result is correct. You can also use a version of Fubini-Tonelli's theorem (i.e. changing order of summation):
$$begin{eqnarray}
sum_{j=0}^infty j(1-p)^{j-1}&=&sum_{j=0}^infty left(sum_{k=0}^{j-1} 1right)(1-p)^{j-1}\&=&sum_{k=0}^infty left(sum_{j=k+1}^infty(1-p)^{j-1} right)\
&=&sum_{k=0}^infty frac{(1-p)^k}{p}\
&=&frac{1}{p^2}.
end{eqnarray}$$ This gives
$$
Bbb E[X]=psum_{j=0}^infty j(1-p)^{j-1}=frac{1}{p}.
$$
Note: Note that $$binom{-2}{j}=frac{(-2)(-3)cdots(-1-j)}{j!}=(-1)^j(j+1).$$ Generalized binomial theorem also gives
$$
sum_{j=0}^infty (j+1)x^j=sum_{j=0}^infty binom{-2}{j}(-x)^j=(1-x)^{-2}.
$$
answered Jan 22 at 21:24
SongSong
17k21145
edited Jan 22 at 21:31
answered Jan 22 at 21:24
SongSong
17k21145
answered Jan 22 at 21:24
SongSong
17k21145
answered Jan 22 at 21:24
SongSong
17k21145
17k21145
add a comment |
add a comment |
$begingroup$
I can offer a slightly different way of doing this
$$ E(X) = p sum_{x=1}^{infty} x(1-p)^{x-1} $$
$$ (1-p)E(X) =psum_{x=1}^{infty} x(1-p)^{x} $$
$$ E(X) - (1-p)E(X) = psum_{x=0}^{infty} (1-p)^{x} $$
You can see this writing down the first few terms in each sum :
$$ E(X) = p( 1 + 2(1-p) + 3(1-p)^2 ...)$$
$$ (1-p)E(X) = p( (1-p)+ 2(1-p)^2 + 3(1-p)^3 ...) $$
$$ E(X) - (1-p)E(X) =p (1+ (1-p) + (1-p)^2 + (1-p)^3 ...) $$
$$ E(X) - (1-p)E(X) = p frac{1}{1-(1-p)} $$
$$ E(X) =frac{1}{p}$$
answered Jan 22 at 21:52
Rohan NuckchadyRohan Nuckchady
1111
$endgroup$
add a comment |
$begingroup$
I can offer a slightly different way of doing this
$$ E(X) = p sum_{x=1}^{infty} x(1-p)^{x-1} $$
$$ (1-p)E(X) =psum_{x=1}^{infty} x(1-p)^{x} $$
$$ E(X) - (1-p)E(X) = psum_{x=0}^{infty} (1-p)^{x} $$
You can see this writing down the first few terms in each sum :
$$ E(X) = p( 1 + 2(1-p) + 3(1-p)^2 ...)$$
$$ (1-p)E(X) = p( (1-p)+ 2(1-p)^2 + 3(1-p)^3 ...) $$
$$ E(X) - (1-p)E(X) =p (1+ (1-p) + (1-p)^2 + (1-p)^3 ...) $$
$$ E(X) - (1-p)E(X) = p frac{1}{1-(1-p)} $$
$$ E(X) =frac{1}{p}$$
answered Jan 22 at 21:52
Rohan NuckchadyRohan Nuckchady
1111
$endgroup$
add a comment |
$begingroup$
I can offer a slightly different way of doing this
$$ E(X) = p sum_{x=1}^{infty} x(1-p)^{x-1} $$
$$ (1-p)E(X) =psum_{x=1}^{infty} x(1-p)^{x} $$
$$ E(X) - (1-p)E(X) = psum_{x=0}^{infty} (1-p)^{x} $$
You can see this writing down the first few terms in each sum :
$$ E(X) = p( 1 + 2(1-p) + 3(1-p)^2 ...)$$
$$ (1-p)E(X) = p( (1-p)+ 2(1-p)^2 + 3(1-p)^3 ...) $$
$$ E(X) - (1-p)E(X) =p (1+ (1-p) + (1-p)^2 + (1-p)^3 ...) $$
$$ E(X) - (1-p)E(X) = p frac{1}{1-(1-p)} $$
$$ E(X) =frac{1}{p}$$
answered Jan 22 at 21:52
Rohan NuckchadyRohan Nuckchady
1111
$endgroup$
I can offer a slightly different way of doing this
$$ E(X) = p sum_{x=1}^{infty} x(1-p)^{x-1} $$
$$ (1-p)E(X) =psum_{x=1}^{infty} x(1-p)^{x} $$
$$ E(X) - (1-p)E(X) = psum_{x=0}^{infty} (1-p)^{x} $$
You can see this writing down the first few terms in each sum :
$$ E(X) = p( 1 + 2(1-p) + 3(1-p)^2 ...)$$
$$ (1-p)E(X) = p( (1-p)+ 2(1-p)^2 + 3(1-p)^3 ...) $$
$$ E(X) - (1-p)E(X) =p (1+ (1-p) + (1-p)^2 + (1-p)^3 ...) $$
$$ E(X) - (1-p)E(X) = p frac{1}{1-(1-p)} $$
$$ E(X) =frac{1}{p}$$
answered Jan 22 at 21:52
Rohan NuckchadyRohan Nuckchady
1111
answered Jan 22 at 21:52
Rohan NuckchadyRohan Nuckchady
1111
answered Jan 22 at 21:52
Rohan NuckchadyRohan Nuckchady
1111
answered Jan 22 at 21:52
Rohan NuckchadyRohan Nuckchady
1111
1111
add a comment |
add a comment |
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