Find the CDF of a given pmf (discrete case)
$begingroup$
I am trying to simulate n random discrete variable which has the following
pmf
$P(X = k) = (1-p)^2kp^{k-1}$
I am thinking about using the inverse transform sampling method and I am trying to find the cdf.
$P(X le k) = 1 - P(X > k) = 1- P(X ge k+1) = sum_{x=k+1}^{infty} (1-p)^2kp^{k-1} = 1-[ (1-p)^2sum_{x=k+1}^{infty}kp^{k-1}]$
= $1-(1-p)^2[(k+1)p^k + (k+2)p^{k+1}+ (k+3)p^{k+4} + ...] $
I can't seem to calculate this sum ($sum_{x=k+1}^{infty}kp^{k-1}$).
Any help or hint will be appreciated !
calculus statistics
asked Dec 3 '18 at 5:39
TatariaTataria
277
$endgroup$
add a comment |
$begingroup$
I am trying to simulate n random discrete variable which has the following
pmf
$P(X = k) = (1-p)^2kp^{k-1}$
I am thinking about using the inverse transform sampling method and I am trying to find the cdf.
$P(X le k) = 1 - P(X > k) = 1- P(X ge k+1) = sum_{x=k+1}^{infty} (1-p)^2kp^{k-1} = 1-[ (1-p)^2sum_{x=k+1}^{infty}kp^{k-1}]$
= $1-(1-p)^2[(k+1)p^k + (k+2)p^{k+1}+ (k+3)p^{k+4} + ...] $
I can't seem to calculate this sum ($sum_{x=k+1}^{infty}kp^{k-1}$).
Any help or hint will be appreciated !
calculus statistics
asked Dec 3 '18 at 5:39
TatariaTataria
277
$endgroup$
add a comment |
$begingroup$
I am trying to simulate n random discrete variable which has the following
pmf
$P(X = k) = (1-p)^2kp^{k-1}$
I am thinking about using the inverse transform sampling method and I am trying to find the cdf.
$P(X le k) = 1 - P(X > k) = 1- P(X ge k+1) = sum_{x=k+1}^{infty} (1-p)^2kp^{k-1} = 1-[ (1-p)^2sum_{x=k+1}^{infty}kp^{k-1}]$
= $1-(1-p)^2[(k+1)p^k + (k+2)p^{k+1}+ (k+3)p^{k+4} + ...] $
I can't seem to calculate this sum ($sum_{x=k+1}^{infty}kp^{k-1}$).
Any help or hint will be appreciated !
calculus statistics
asked Dec 3 '18 at 5:39
TatariaTataria
277
$endgroup$
I am trying to simulate n random discrete variable which has the following
pmf
$P(X = k) = (1-p)^2kp^{k-1}$
I am thinking about using the inverse transform sampling method and I am trying to find the cdf.
$P(X le k) = 1 - P(X > k) = 1- P(X ge k+1) = sum_{x=k+1}^{infty} (1-p)^2kp^{k-1} = 1-[ (1-p)^2sum_{x=k+1}^{infty}kp^{k-1}]$
= $1-(1-p)^2[(k+1)p^k + (k+2)p^{k+1}+ (k+3)p^{k+4} + ...] $
I can't seem to calculate this sum ($sum_{x=k+1}^{infty}kp^{k-1}$).
Any help or hint will be appreciated !
calculus statistics
calculus statistics
asked Dec 3 '18 at 5:39
TatariaTataria
277
asked Dec 3 '18 at 5:39
TatariaTataria
277
asked Dec 3 '18 at 5:39
TatariaTataria
277
asked Dec 3 '18 at 5:39
TatariaTataria
277
asked Dec 3 '18 at 5:39
TatariaTataria
277
277
add a comment |
add a comment |
1 Answer
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$begingroup$
begin{align}S &= (k+1) p^k + (k+2) p^{k+1} + (k+3) p^{k+2} + cdots
\
pS &= phantom{(k+1) p^k +{}} (k+1) p^{k+1} + (k+2) p^{k+2} + (k+3) p^{k+2} + cdots
end{align}
Subtracting the two equations yields
$$(1-p) S = p^k + p^{k+1} + p^{k+2} + cdots.$$
The right-hand side is a geometric series I assume you can compute. Then divide both sides by $1-p$.
answered Dec 3 '18 at 5:47
angryavianangryavian
40.6k23280
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1 Answer
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$begingroup$
begin{align}S &= (k+1) p^k + (k+2) p^{k+1} + (k+3) p^{k+2} + cdots
\
pS &= phantom{(k+1) p^k +{}} (k+1) p^{k+1} + (k+2) p^{k+2} + (k+3) p^{k+2} + cdots
end{align}
Subtracting the two equations yields
$$(1-p) S = p^k + p^{k+1} + p^{k+2} + cdots.$$
The right-hand side is a geometric series I assume you can compute. Then divide both sides by $1-p$.
answered Dec 3 '18 at 5:47
angryavianangryavian
40.6k23280
$endgroup$
add a comment |
$begingroup$
begin{align}S &= (k+1) p^k + (k+2) p^{k+1} + (k+3) p^{k+2} + cdots
\
pS &= phantom{(k+1) p^k +{}} (k+1) p^{k+1} + (k+2) p^{k+2} + (k+3) p^{k+2} + cdots
end{align}
Subtracting the two equations yields
$$(1-p) S = p^k + p^{k+1} + p^{k+2} + cdots.$$
The right-hand side is a geometric series I assume you can compute. Then divide both sides by $1-p$.
answered Dec 3 '18 at 5:47
angryavianangryavian
40.6k23280
$endgroup$
add a comment |
$begingroup$
begin{align}S &= (k+1) p^k + (k+2) p^{k+1} + (k+3) p^{k+2} + cdots
\
pS &= phantom{(k+1) p^k +{}} (k+1) p^{k+1} + (k+2) p^{k+2} + (k+3) p^{k+2} + cdots
end{align}
Subtracting the two equations yields
$$(1-p) S = p^k + p^{k+1} + p^{k+2} + cdots.$$
The right-hand side is a geometric series I assume you can compute. Then divide both sides by $1-p$.
answered Dec 3 '18 at 5:47
angryavianangryavian
40.6k23280
$endgroup$
begin{align}S &= (k+1) p^k + (k+2) p^{k+1} + (k+3) p^{k+2} + cdots
\
pS &= phantom{(k+1) p^k +{}} (k+1) p^{k+1} + (k+2) p^{k+2} + (k+3) p^{k+2} + cdots
end{align}
Subtracting the two equations yields
$$(1-p) S = p^k + p^{k+1} + p^{k+2} + cdots.$$
The right-hand side is a geometric series I assume you can compute. Then divide both sides by $1-p$.
answered Dec 3 '18 at 5:47
angryavianangryavian
40.6k23280
answered Dec 3 '18 at 5:47
angryavianangryavian
40.6k23280
answered Dec 3 '18 at 5:47
angryavianangryavian
40.6k23280
answered Dec 3 '18 at 5:47
angryavianangryavian
40.6k23280
40.6k23280
add a comment |
add a comment |
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