Expected value of first success in Bernoulli trials
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I have a sequence of $n$ Bernoulli experiments each succeeding with probability $p$.
In general, the number of successes $X$ follows the well-known binomial distribution:
$$P(X = k) = begin{pmatrix} n\ k end{pmatrix} p^k(1-p)^{n-k}$$
However, I'm interested in the expected value of the number of experiments until the first success, assuming a maximum number of $n$ trials (if the experiment never succeeds, there will have been $n$ experiments attempted).
My reasoning is the following: if $n = 4$, the expected value will be (?):
$$ 1cdot p + 2cdot(1-p)p + 3cdot(1-p)^2p + 4cdot(1-p)^3p + 5(1-p)^5$$
Generalizing:
$$E = sum_{i=1}^n icdot p(1-p)^{i-1}$$
However, the sum of these weights is not one, as it should be:
$$sum_{i=1}^n p(1-p)^{i-1} neq 1$$
Clearly I'm off somewhere. What am I missing?
probability-distributions binomial-distribution expected-value
$endgroup$
|
show 1 more comment
$begingroup$
I have a sequence of $n$ Bernoulli experiments each succeeding with probability $p$.
In general, the number of successes $X$ follows the well-known binomial distribution:
$$P(X = k) = begin{pmatrix} n\ k end{pmatrix} p^k(1-p)^{n-k}$$
However, I'm interested in the expected value of the number of experiments until the first success, assuming a maximum number of $n$ trials (if the experiment never succeeds, there will have been $n$ experiments attempted).
My reasoning is the following: if $n = 4$, the expected value will be (?):
$$ 1cdot p + 2cdot(1-p)p + 3cdot(1-p)^2p + 4cdot(1-p)^3p + 5(1-p)^5$$
Generalizing:
$$E = sum_{i=1}^n icdot p(1-p)^{i-1}$$
However, the sum of these weights is not one, as it should be:
$$sum_{i=1}^n p(1-p)^{i-1} neq 1$$
Clearly I'm off somewhere. What am I missing?
probability-distributions binomial-distribution expected-value
$endgroup$
$begingroup$
Could you please elaborate on why exactly it should be $1$? Keep in mind that you are calculating expected value of the random variable, not probability distribution.
$endgroup$
– Sauhard Sharma
Dec 13 '18 at 14:15
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The expected value shouldn't be one, but the sum of the coefficients should
$endgroup$
– Sean Bone
Dec 13 '18 at 14:21
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If I understand well then to be found is the expected value of a random variable $Z$. It will take a value in ${1,dots,n}$ if a success appears in the first $n$ trials. Could you tell us what value $Z$ will take if that does not happen?
$endgroup$
– drhab
Dec 13 '18 at 14:28
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That is correct. If none of the $n$ trials is successful, $Z$ will be $n$.
$endgroup$
– Sean Bone
Dec 13 '18 at 14:31
$begingroup$
Your probability distribution is based on the assumption that it will succeed ultimately. If you add in the probability that it doesn't succeed at all, your weights will add up to $1$.
$endgroup$
– Sauhard Sharma
Dec 13 '18 at 14:31
|
show 1 more comment
$begingroup$
I have a sequence of $n$ Bernoulli experiments each succeeding with probability $p$.
In general, the number of successes $X$ follows the well-known binomial distribution:
$$P(X = k) = begin{pmatrix} n\ k end{pmatrix} p^k(1-p)^{n-k}$$
However, I'm interested in the expected value of the number of experiments until the first success, assuming a maximum number of $n$ trials (if the experiment never succeeds, there will have been $n$ experiments attempted).
My reasoning is the following: if $n = 4$, the expected value will be (?):
$$ 1cdot p + 2cdot(1-p)p + 3cdot(1-p)^2p + 4cdot(1-p)^3p + 5(1-p)^5$$
Generalizing:
$$E = sum_{i=1}^n icdot p(1-p)^{i-1}$$
However, the sum of these weights is not one, as it should be:
$$sum_{i=1}^n p(1-p)^{i-1} neq 1$$
Clearly I'm off somewhere. What am I missing?
probability-distributions binomial-distribution expected-value
$endgroup$
I have a sequence of $n$ Bernoulli experiments each succeeding with probability $p$.
In general, the number of successes $X$ follows the well-known binomial distribution:
$$P(X = k) = begin{pmatrix} n\ k end{pmatrix} p^k(1-p)^{n-k}$$
However, I'm interested in the expected value of the number of experiments until the first success, assuming a maximum number of $n$ trials (if the experiment never succeeds, there will have been $n$ experiments attempted).
My reasoning is the following: if $n = 4$, the expected value will be (?):
$$ 1cdot p + 2cdot(1-p)p + 3cdot(1-p)^2p + 4cdot(1-p)^3p + 5(1-p)^5$$
Generalizing:
$$E = sum_{i=1}^n icdot p(1-p)^{i-1}$$
However, the sum of these weights is not one, as it should be:
$$sum_{i=1}^n p(1-p)^{i-1} neq 1$$
Clearly I'm off somewhere. What am I missing?
probability-distributions binomial-distribution expected-value
probability-distributions binomial-distribution expected-value
edited Dec 13 '18 at 14:21
Sean Bone
asked Dec 13 '18 at 14:11
Sean BoneSean Bone
1345
1345
$begingroup$
Could you please elaborate on why exactly it should be $1$? Keep in mind that you are calculating expected value of the random variable, not probability distribution.
$endgroup$
– Sauhard Sharma
Dec 13 '18 at 14:15
$begingroup$
The expected value shouldn't be one, but the sum of the coefficients should
$endgroup$
– Sean Bone
Dec 13 '18 at 14:21
$begingroup$
If I understand well then to be found is the expected value of a random variable $Z$. It will take a value in ${1,dots,n}$ if a success appears in the first $n$ trials. Could you tell us what value $Z$ will take if that does not happen?
$endgroup$
– drhab
Dec 13 '18 at 14:28
$begingroup$
That is correct. If none of the $n$ trials is successful, $Z$ will be $n$.
$endgroup$
– Sean Bone
Dec 13 '18 at 14:31
$begingroup$
Your probability distribution is based on the assumption that it will succeed ultimately. If you add in the probability that it doesn't succeed at all, your weights will add up to $1$.
$endgroup$
– Sauhard Sharma
Dec 13 '18 at 14:31
|
show 1 more comment
$begingroup$
Could you please elaborate on why exactly it should be $1$? Keep in mind that you are calculating expected value of the random variable, not probability distribution.
$endgroup$
– Sauhard Sharma
Dec 13 '18 at 14:15
$begingroup$
The expected value shouldn't be one, but the sum of the coefficients should
$endgroup$
– Sean Bone
Dec 13 '18 at 14:21
$begingroup$
If I understand well then to be found is the expected value of a random variable $Z$. It will take a value in ${1,dots,n}$ if a success appears in the first $n$ trials. Could you tell us what value $Z$ will take if that does not happen?
$endgroup$
– drhab
Dec 13 '18 at 14:28
$begingroup$
That is correct. If none of the $n$ trials is successful, $Z$ will be $n$.
$endgroup$
– Sean Bone
Dec 13 '18 at 14:31
$begingroup$
Your probability distribution is based on the assumption that it will succeed ultimately. If you add in the probability that it doesn't succeed at all, your weights will add up to $1$.
$endgroup$
– Sauhard Sharma
Dec 13 '18 at 14:31
$begingroup$
Could you please elaborate on why exactly it should be $1$? Keep in mind that you are calculating expected value of the random variable, not probability distribution.
$endgroup$
– Sauhard Sharma
Dec 13 '18 at 14:15
$begingroup$
Could you please elaborate on why exactly it should be $1$? Keep in mind that you are calculating expected value of the random variable, not probability distribution.
$endgroup$
– Sauhard Sharma
Dec 13 '18 at 14:15
$begingroup$
The expected value shouldn't be one, but the sum of the coefficients should
$endgroup$
– Sean Bone
Dec 13 '18 at 14:21
$begingroup$
The expected value shouldn't be one, but the sum of the coefficients should
$endgroup$
– Sean Bone
Dec 13 '18 at 14:21
$begingroup$
If I understand well then to be found is the expected value of a random variable $Z$. It will take a value in ${1,dots,n}$ if a success appears in the first $n$ trials. Could you tell us what value $Z$ will take if that does not happen?
$endgroup$
– drhab
Dec 13 '18 at 14:28
$begingroup$
If I understand well then to be found is the expected value of a random variable $Z$. It will take a value in ${1,dots,n}$ if a success appears in the first $n$ trials. Could you tell us what value $Z$ will take if that does not happen?
$endgroup$
– drhab
Dec 13 '18 at 14:28
$begingroup$
That is correct. If none of the $n$ trials is successful, $Z$ will be $n$.
$endgroup$
– Sean Bone
Dec 13 '18 at 14:31
$begingroup$
That is correct. If none of the $n$ trials is successful, $Z$ will be $n$.
$endgroup$
– Sean Bone
Dec 13 '18 at 14:31
$begingroup$
Your probability distribution is based on the assumption that it will succeed ultimately. If you add in the probability that it doesn't succeed at all, your weights will add up to $1$.
$endgroup$
– Sauhard Sharma
Dec 13 '18 at 14:31
$begingroup$
Your probability distribution is based on the assumption that it will succeed ultimately. If you add in the probability that it doesn't succeed at all, your weights will add up to $1$.
$endgroup$
– Sauhard Sharma
Dec 13 '18 at 14:31
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
Expected value:
$$E = ncdot(1 - sum_{i=1}^N p(1-p)^{i-1}) + sum_{i=1}^N icdot p(1-p)^{i-1}$$
$endgroup$
$begingroup$
As it hasn't been mentioned yet in comments, in case you don't know, your scenario is similar to a Geometric distribution.
$endgroup$
– Sambo
Dec 13 '18 at 14:39
$begingroup$
I think $N$ should be changed into $n-1$
$endgroup$
– drhab
Dec 13 '18 at 15:20
add a comment |
$begingroup$
Denoting $q:=1-p$ for $n=4$ we find:
$$pcdot1+qpcdot2+q^2pcdot3+(1-p-qp-q^2p)cdot4$$
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Expected value:
$$E = ncdot(1 - sum_{i=1}^N p(1-p)^{i-1}) + sum_{i=1}^N icdot p(1-p)^{i-1}$$
$endgroup$
$begingroup$
As it hasn't been mentioned yet in comments, in case you don't know, your scenario is similar to a Geometric distribution.
$endgroup$
– Sambo
Dec 13 '18 at 14:39
$begingroup$
I think $N$ should be changed into $n-1$
$endgroup$
– drhab
Dec 13 '18 at 15:20
add a comment |
$begingroup$
Expected value:
$$E = ncdot(1 - sum_{i=1}^N p(1-p)^{i-1}) + sum_{i=1}^N icdot p(1-p)^{i-1}$$
$endgroup$
$begingroup$
As it hasn't been mentioned yet in comments, in case you don't know, your scenario is similar to a Geometric distribution.
$endgroup$
– Sambo
Dec 13 '18 at 14:39
$begingroup$
I think $N$ should be changed into $n-1$
$endgroup$
– drhab
Dec 13 '18 at 15:20
add a comment |
$begingroup$
Expected value:
$$E = ncdot(1 - sum_{i=1}^N p(1-p)^{i-1}) + sum_{i=1}^N icdot p(1-p)^{i-1}$$
$endgroup$
Expected value:
$$E = ncdot(1 - sum_{i=1}^N p(1-p)^{i-1}) + sum_{i=1}^N icdot p(1-p)^{i-1}$$
answered Dec 13 '18 at 14:34
Sean BoneSean Bone
1345
1345
$begingroup$
As it hasn't been mentioned yet in comments, in case you don't know, your scenario is similar to a Geometric distribution.
$endgroup$
– Sambo
Dec 13 '18 at 14:39
$begingroup$
I think $N$ should be changed into $n-1$
$endgroup$
– drhab
Dec 13 '18 at 15:20
add a comment |
$begingroup$
As it hasn't been mentioned yet in comments, in case you don't know, your scenario is similar to a Geometric distribution.
$endgroup$
– Sambo
Dec 13 '18 at 14:39
$begingroup$
I think $N$ should be changed into $n-1$
$endgroup$
– drhab
Dec 13 '18 at 15:20
$begingroup$
As it hasn't been mentioned yet in comments, in case you don't know, your scenario is similar to a Geometric distribution.
$endgroup$
– Sambo
Dec 13 '18 at 14:39
$begingroup$
As it hasn't been mentioned yet in comments, in case you don't know, your scenario is similar to a Geometric distribution.
$endgroup$
– Sambo
Dec 13 '18 at 14:39
$begingroup$
I think $N$ should be changed into $n-1$
$endgroup$
– drhab
Dec 13 '18 at 15:20
$begingroup$
I think $N$ should be changed into $n-1$
$endgroup$
– drhab
Dec 13 '18 at 15:20
add a comment |
$begingroup$
Denoting $q:=1-p$ for $n=4$ we find:
$$pcdot1+qpcdot2+q^2pcdot3+(1-p-qp-q^2p)cdot4$$
$endgroup$
add a comment |
$begingroup$
Denoting $q:=1-p$ for $n=4$ we find:
$$pcdot1+qpcdot2+q^2pcdot3+(1-p-qp-q^2p)cdot4$$
$endgroup$
add a comment |
$begingroup$
Denoting $q:=1-p$ for $n=4$ we find:
$$pcdot1+qpcdot2+q^2pcdot3+(1-p-qp-q^2p)cdot4$$
$endgroup$
Denoting $q:=1-p$ for $n=4$ we find:
$$pcdot1+qpcdot2+q^2pcdot3+(1-p-qp-q^2p)cdot4$$
answered Dec 13 '18 at 15:17
drhabdrhab
101k544130
101k544130
add a comment |
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$begingroup$
Could you please elaborate on why exactly it should be $1$? Keep in mind that you are calculating expected value of the random variable, not probability distribution.
$endgroup$
– Sauhard Sharma
Dec 13 '18 at 14:15
$begingroup$
The expected value shouldn't be one, but the sum of the coefficients should
$endgroup$
– Sean Bone
Dec 13 '18 at 14:21
$begingroup$
If I understand well then to be found is the expected value of a random variable $Z$. It will take a value in ${1,dots,n}$ if a success appears in the first $n$ trials. Could you tell us what value $Z$ will take if that does not happen?
$endgroup$
– drhab
Dec 13 '18 at 14:28
$begingroup$
That is correct. If none of the $n$ trials is successful, $Z$ will be $n$.
$endgroup$
– Sean Bone
Dec 13 '18 at 14:31
$begingroup$
Your probability distribution is based on the assumption that it will succeed ultimately. If you add in the probability that it doesn't succeed at all, your weights will add up to $1$.
$endgroup$
– Sauhard Sharma
Dec 13 '18 at 14:31