Deriving likelihood function of binomial distribution, confusion over exponents
$begingroup$
This question focuses on a specific aspect of this one:
How to derive the likelihood function for binomial distribution for parameter estimation?
In my own derivation, I start with:
$$f(xmid p) = mC_x~p^x(1-p)^{m-x}$$
Ignoring $mC_x$, the likelihood function is then given by:
$$L(p) = prod_{i=1}^np^{x_i}(1-p)^{m-x_i} = p^{sum_1^n x_i}(1-p)^{sum_1^n m-x_i} = p^{x}(1-p)^{nm-x}$$
However, in the question I referenced, they have this instead:
$$prod_{i=1}^np^{x_i}(1-p)^{1-x_i} = p^{sum_1^n x_i}(1-p)^{sum_1^n1-x_i} = p^{x}(1-p)^{n-x}$$
My question is, are both approaches correct? If so, why does the referenced question use $1$ in place of $m$ on the exponents?
estimation maximum-likelihood binomial likelihood point-estimation
edited Dec 23 '18 at 18:09
Michael Hardy
3,8651430
asked Dec 23 '18 at 17:11
HumptyDumptyHumptyDumpty
1183
$endgroup$
add a comment |
$begingroup$
This question focuses on a specific aspect of this one:
How to derive the likelihood function for binomial distribution for parameter estimation?
In my own derivation, I start with:
$$f(xmid p) = mC_x~p^x(1-p)^{m-x}$$
Ignoring $mC_x$, the likelihood function is then given by:
$$L(p) = prod_{i=1}^np^{x_i}(1-p)^{m-x_i} = p^{sum_1^n x_i}(1-p)^{sum_1^n m-x_i} = p^{x}(1-p)^{nm-x}$$
However, in the question I referenced, they have this instead:
$$prod_{i=1}^np^{x_i}(1-p)^{1-x_i} = p^{sum_1^n x_i}(1-p)^{sum_1^n1-x_i} = p^{x}(1-p)^{n-x}$$
My question is, are both approaches correct? If so, why does the referenced question use $1$ in place of $m$ on the exponents?
estimation maximum-likelihood binomial likelihood point-estimation
edited Dec 23 '18 at 18:09
Michael Hardy
3,8651430
asked Dec 23 '18 at 17:11
HumptyDumptyHumptyDumpty
1183
$endgroup$
add a comment |
$begingroup$
This question focuses on a specific aspect of this one:
How to derive the likelihood function for binomial distribution for parameter estimation?
In my own derivation, I start with:
$$f(xmid p) = mC_x~p^x(1-p)^{m-x}$$
Ignoring $mC_x$, the likelihood function is then given by:
$$L(p) = prod_{i=1}^np^{x_i}(1-p)^{m-x_i} = p^{sum_1^n x_i}(1-p)^{sum_1^n m-x_i} = p^{x}(1-p)^{nm-x}$$
However, in the question I referenced, they have this instead:
$$prod_{i=1}^np^{x_i}(1-p)^{1-x_i} = p^{sum_1^n x_i}(1-p)^{sum_1^n1-x_i} = p^{x}(1-p)^{n-x}$$
My question is, are both approaches correct? If so, why does the referenced question use $1$ in place of $m$ on the exponents?
estimation maximum-likelihood binomial likelihood point-estimation
edited Dec 23 '18 at 18:09
Michael Hardy
3,8651430
asked Dec 23 '18 at 17:11
HumptyDumptyHumptyDumpty
1183
$endgroup$
This question focuses on a specific aspect of this one:
How to derive the likelihood function for binomial distribution for parameter estimation?
In my own derivation, I start with:
$$f(xmid p) = mC_x~p^x(1-p)^{m-x}$$
Ignoring $mC_x$, the likelihood function is then given by:
$$L(p) = prod_{i=1}^np^{x_i}(1-p)^{m-x_i} = p^{sum_1^n x_i}(1-p)^{sum_1^n m-x_i} = p^{x}(1-p)^{nm-x}$$
However, in the question I referenced, they have this instead:
$$prod_{i=1}^np^{x_i}(1-p)^{1-x_i} = p^{sum_1^n x_i}(1-p)^{sum_1^n1-x_i} = p^{x}(1-p)^{n-x}$$
My question is, are both approaches correct? If so, why does the referenced question use $1$ in place of $m$ on the exponents?
estimation maximum-likelihood binomial likelihood point-estimation
estimation maximum-likelihood binomial likelihood point-estimation
edited Dec 23 '18 at 18:09
Michael Hardy
3,8651430
asked Dec 23 '18 at 17:11
HumptyDumptyHumptyDumpty
1183
edited Dec 23 '18 at 18:09
Michael Hardy
3,8651430
asked Dec 23 '18 at 17:11
HumptyDumptyHumptyDumpty
1183
edited Dec 23 '18 at 18:09
Michael Hardy
3,8651430
edited Dec 23 '18 at 18:09
Michael Hardy
3,8651430
edited Dec 23 '18 at 18:09
Michael Hardy
3,8651430
3,8651430
asked Dec 23 '18 at 17:11
HumptyDumptyHumptyDumpty
1183
asked Dec 23 '18 at 17:11
HumptyDumptyHumptyDumpty
1183
asked Dec 23 '18 at 17:11
HumptyDumptyHumptyDumpty
1183
1183
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
It looks as if you intended $X_1,ldots,X_n sim operatorname{i{.}i{.}d{.}} operatorname{Binomial}(m,p).$ Then you have
$$
L(p) propto prod_{i=1}^n p^{x_i} (1-p)^{m-x_i} = p^{sum_{i=1}^n x_i} (1-p)^{nm - sum_{i=1}^n x_i} = p^x (1-p)^{nm-x}.
$$
It appears that in the question that you "referenced" (I don't know what "referenced" means in this context, but that doesn't appear to matter.) one has
$X_1,ldots,X_n sim operatorname{i{.}i{.}d{.}} operatorname{Bernoulli}(p)$, so that $X_1+cdots+X_n sim operatorname{Binomial} (n,p).$ That leads to
$$
L(p) propto p^{sum_{i=1}^n x_i} (1-p)^{n - sum_{i=1}^n x_i} = p^x(1-p)^{n-x}.
$$
Therefore both are right, but they're answers to different questions.
answered Dec 23 '18 at 18:17
Michael HardyMichael Hardy
3,8651430
$endgroup$
add a comment |
$begingroup$
The question you are referencing is starting with a Bernoulli distribution. To be sure, $x$ only takes on 0 or 1 in that question. In your work, you are starting with a Binomial distribution. To be sure, your values of $x=0, 1, 2, ... m$.
Remember that the sum of $n$ independent $Bernoulli(p)$ variables is a $Binomial(n, p)$ distribution. This should account for the differences you are seeing.
Your derivation for the likelihood of a binomial is just fine, ignoring the $mCx$ term, but you shouldn't ignore it. You can treat it as ignorable for the purposes of calculating the likelihood function since the likelihood function is a function only of the parameter $p$ and $p$ does not show up in $mCx$.
answered Dec 23 '18 at 17:36
StatsStudentStatsStudent
6,01332044
$endgroup$
$begingroup$
Are you using "To be sure, [...]" to mean "You can be sure of this because [...]"? (I ask because "To be sure, [...]" means something quite different in English, which makes your usage very confusing. If you didn't intend the idiomatic meaning, I recommend finding a different phrase to express what you meant.)
$endgroup$
– ruakh
Dec 24 '18 at 4:07
$begingroup$
@ruakh, indeed the intention was idiomatic. It is correctly used to mean "certainly, undoubtedly, admittedly." I will keep it.
$endgroup$
– StatsStudent
Dec 26 '18 at 17:27
$begingroup$
But "admittedly" makes no sense in this context. You must be misunderstanding the idiom. :-/
$endgroup$
– ruakh
Dec 26 '18 at 18:02
$begingroup$
No, I have a perfect understanding. I'm using the idiomatic meaning of "certainly" or "undoubtedly" - not "admittedly."
$endgroup$
– StatsStudent
Dec 26 '18 at 18:03
$begingroup$
That's not a separate meaning. You can use "certainly" to mean "to be sure" (for example, "To be sure, not everyone does it; but most people do" can be rephrased as "Certainly, not everyone does it; but most people do"); but you can't use "to be sure" in all cases where you can use "certainly" (for example, "It's certainly possible; would you like us to do it now?" can't be rephrased as "To be sure, it's possible; would you like us to do it now?"). If the word "admittedly" doesn't work, then the phrase "to be sure" doesn't work, either.
$endgroup$
– ruakh
Dec 26 '18 at 18:38
|
show 2 more comments
$begingroup$
The PMF in the question, $f(x_i)=p^{x_i}(1-p)^{1-x_i}$ belongs to Bernoulli distribution, where $x_i$ is a binary variable. Yours is the PMF of Binomial distribution, and $x_i$'s are Binomial RVs ($n$ of them actually) with parameters $(m,p)$.
answered Dec 23 '18 at 17:39
gunesgunes
5,7601115
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It looks as if you intended $X_1,ldots,X_n sim operatorname{i{.}i{.}d{.}} operatorname{Binomial}(m,p).$ Then you have
$$
L(p) propto prod_{i=1}^n p^{x_i} (1-p)^{m-x_i} = p^{sum_{i=1}^n x_i} (1-p)^{nm - sum_{i=1}^n x_i} = p^x (1-p)^{nm-x}.
$$
It appears that in the question that you "referenced" (I don't know what "referenced" means in this context, but that doesn't appear to matter.) one has
$X_1,ldots,X_n sim operatorname{i{.}i{.}d{.}} operatorname{Bernoulli}(p)$, so that $X_1+cdots+X_n sim operatorname{Binomial} (n,p).$ That leads to
$$
L(p) propto p^{sum_{i=1}^n x_i} (1-p)^{n - sum_{i=1}^n x_i} = p^x(1-p)^{n-x}.
$$
Therefore both are right, but they're answers to different questions.
answered Dec 23 '18 at 18:17
Michael HardyMichael Hardy
3,8651430
$endgroup$
add a comment |
$begingroup$
It looks as if you intended $X_1,ldots,X_n sim operatorname{i{.}i{.}d{.}} operatorname{Binomial}(m,p).$ Then you have
$$
L(p) propto prod_{i=1}^n p^{x_i} (1-p)^{m-x_i} = p^{sum_{i=1}^n x_i} (1-p)^{nm - sum_{i=1}^n x_i} = p^x (1-p)^{nm-x}.
$$
It appears that in the question that you "referenced" (I don't know what "referenced" means in this context, but that doesn't appear to matter.) one has
$X_1,ldots,X_n sim operatorname{i{.}i{.}d{.}} operatorname{Bernoulli}(p)$, so that $X_1+cdots+X_n sim operatorname{Binomial} (n,p).$ That leads to
$$
L(p) propto p^{sum_{i=1}^n x_i} (1-p)^{n - sum_{i=1}^n x_i} = p^x(1-p)^{n-x}.
$$
Therefore both are right, but they're answers to different questions.
answered Dec 23 '18 at 18:17
Michael HardyMichael Hardy
3,8651430
$endgroup$
add a comment |
$begingroup$
It looks as if you intended $X_1,ldots,X_n sim operatorname{i{.}i{.}d{.}} operatorname{Binomial}(m,p).$ Then you have
$$
L(p) propto prod_{i=1}^n p^{x_i} (1-p)^{m-x_i} = p^{sum_{i=1}^n x_i} (1-p)^{nm - sum_{i=1}^n x_i} = p^x (1-p)^{nm-x}.
$$
It appears that in the question that you "referenced" (I don't know what "referenced" means in this context, but that doesn't appear to matter.) one has
$X_1,ldots,X_n sim operatorname{i{.}i{.}d{.}} operatorname{Bernoulli}(p)$, so that $X_1+cdots+X_n sim operatorname{Binomial} (n,p).$ That leads to
$$
L(p) propto p^{sum_{i=1}^n x_i} (1-p)^{n - sum_{i=1}^n x_i} = p^x(1-p)^{n-x}.
$$
Therefore both are right, but they're answers to different questions.
answered Dec 23 '18 at 18:17
Michael HardyMichael Hardy
3,8651430
$endgroup$
It looks as if you intended $X_1,ldots,X_n sim operatorname{i{.}i{.}d{.}} operatorname{Binomial}(m,p).$ Then you have
$$
L(p) propto prod_{i=1}^n p^{x_i} (1-p)^{m-x_i} = p^{sum_{i=1}^n x_i} (1-p)^{nm - sum_{i=1}^n x_i} = p^x (1-p)^{nm-x}.
$$
It appears that in the question that you "referenced" (I don't know what "referenced" means in this context, but that doesn't appear to matter.) one has
$X_1,ldots,X_n sim operatorname{i{.}i{.}d{.}} operatorname{Bernoulli}(p)$, so that $X_1+cdots+X_n sim operatorname{Binomial} (n,p).$ That leads to
$$
L(p) propto p^{sum_{i=1}^n x_i} (1-p)^{n - sum_{i=1}^n x_i} = p^x(1-p)^{n-x}.
$$
Therefore both are right, but they're answers to different questions.
answered Dec 23 '18 at 18:17
Michael HardyMichael Hardy
3,8651430
answered Dec 23 '18 at 18:17
Michael HardyMichael Hardy
3,8651430
answered Dec 23 '18 at 18:17
Michael HardyMichael Hardy
3,8651430
answered Dec 23 '18 at 18:17
Michael HardyMichael Hardy
3,8651430
3,8651430
add a comment |
add a comment |
$begingroup$
The question you are referencing is starting with a Bernoulli distribution. To be sure, $x$ only takes on 0 or 1 in that question. In your work, you are starting with a Binomial distribution. To be sure, your values of $x=0, 1, 2, ... m$.
Remember that the sum of $n$ independent $Bernoulli(p)$ variables is a $Binomial(n, p)$ distribution. This should account for the differences you are seeing.
Your derivation for the likelihood of a binomial is just fine, ignoring the $mCx$ term, but you shouldn't ignore it. You can treat it as ignorable for the purposes of calculating the likelihood function since the likelihood function is a function only of the parameter $p$ and $p$ does not show up in $mCx$.
answered Dec 23 '18 at 17:36
StatsStudentStatsStudent
6,01332044
$endgroup$
$begingroup$
Are you using "To be sure, [...]" to mean "You can be sure of this because [...]"? (I ask because "To be sure, [...]" means something quite different in English, which makes your usage very confusing. If you didn't intend the idiomatic meaning, I recommend finding a different phrase to express what you meant.)
$endgroup$
– ruakh
Dec 24 '18 at 4:07
$begingroup$
@ruakh, indeed the intention was idiomatic. It is correctly used to mean "certainly, undoubtedly, admittedly." I will keep it.
$endgroup$
– StatsStudent
Dec 26 '18 at 17:27
$begingroup$
But "admittedly" makes no sense in this context. You must be misunderstanding the idiom. :-/
$endgroup$
– ruakh
Dec 26 '18 at 18:02
$begingroup$
No, I have a perfect understanding. I'm using the idiomatic meaning of "certainly" or "undoubtedly" - not "admittedly."
$endgroup$
– StatsStudent
Dec 26 '18 at 18:03
$begingroup$
That's not a separate meaning. You can use "certainly" to mean "to be sure" (for example, "To be sure, not everyone does it; but most people do" can be rephrased as "Certainly, not everyone does it; but most people do"); but you can't use "to be sure" in all cases where you can use "certainly" (for example, "It's certainly possible; would you like us to do it now?" can't be rephrased as "To be sure, it's possible; would you like us to do it now?"). If the word "admittedly" doesn't work, then the phrase "to be sure" doesn't work, either.
$endgroup$
– ruakh
Dec 26 '18 at 18:38
|
show 2 more comments
$begingroup$
The question you are referencing is starting with a Bernoulli distribution. To be sure, $x$ only takes on 0 or 1 in that question. In your work, you are starting with a Binomial distribution. To be sure, your values of $x=0, 1, 2, ... m$.
Remember that the sum of $n$ independent $Bernoulli(p)$ variables is a $Binomial(n, p)$ distribution. This should account for the differences you are seeing.
Your derivation for the likelihood of a binomial is just fine, ignoring the $mCx$ term, but you shouldn't ignore it. You can treat it as ignorable for the purposes of calculating the likelihood function since the likelihood function is a function only of the parameter $p$ and $p$ does not show up in $mCx$.
answered Dec 23 '18 at 17:36
StatsStudentStatsStudent
6,01332044
$endgroup$
$begingroup$
Are you using "To be sure, [...]" to mean "You can be sure of this because [...]"? (I ask because "To be sure, [...]" means something quite different in English, which makes your usage very confusing. If you didn't intend the idiomatic meaning, I recommend finding a different phrase to express what you meant.)
$endgroup$
– ruakh
Dec 24 '18 at 4:07
$begingroup$
@ruakh, indeed the intention was idiomatic. It is correctly used to mean "certainly, undoubtedly, admittedly." I will keep it.
$endgroup$
– StatsStudent
Dec 26 '18 at 17:27
$begingroup$
But "admittedly" makes no sense in this context. You must be misunderstanding the idiom. :-/
$endgroup$
– ruakh
Dec 26 '18 at 18:02
$begingroup$
No, I have a perfect understanding. I'm using the idiomatic meaning of "certainly" or "undoubtedly" - not "admittedly."
$endgroup$
– StatsStudent
Dec 26 '18 at 18:03
$begingroup$
That's not a separate meaning. You can use "certainly" to mean "to be sure" (for example, "To be sure, not everyone does it; but most people do" can be rephrased as "Certainly, not everyone does it; but most people do"); but you can't use "to be sure" in all cases where you can use "certainly" (for example, "It's certainly possible; would you like us to do it now?" can't be rephrased as "To be sure, it's possible; would you like us to do it now?"). If the word "admittedly" doesn't work, then the phrase "to be sure" doesn't work, either.
$endgroup$
– ruakh
Dec 26 '18 at 18:38
|
show 2 more comments
$begingroup$
The question you are referencing is starting with a Bernoulli distribution. To be sure, $x$ only takes on 0 or 1 in that question. In your work, you are starting with a Binomial distribution. To be sure, your values of $x=0, 1, 2, ... m$.
Remember that the sum of $n$ independent $Bernoulli(p)$ variables is a $Binomial(n, p)$ distribution. This should account for the differences you are seeing.
Your derivation for the likelihood of a binomial is just fine, ignoring the $mCx$ term, but you shouldn't ignore it. You can treat it as ignorable for the purposes of calculating the likelihood function since the likelihood function is a function only of the parameter $p$ and $p$ does not show up in $mCx$.
answered Dec 23 '18 at 17:36
StatsStudentStatsStudent
6,01332044
$endgroup$
The question you are referencing is starting with a Bernoulli distribution. To be sure, $x$ only takes on 0 or 1 in that question. In your work, you are starting with a Binomial distribution. To be sure, your values of $x=0, 1, 2, ... m$.
Remember that the sum of $n$ independent $Bernoulli(p)$ variables is a $Binomial(n, p)$ distribution. This should account for the differences you are seeing.
Your derivation for the likelihood of a binomial is just fine, ignoring the $mCx$ term, but you shouldn't ignore it. You can treat it as ignorable for the purposes of calculating the likelihood function since the likelihood function is a function only of the parameter $p$ and $p$ does not show up in $mCx$.
answered Dec 23 '18 at 17:36
StatsStudentStatsStudent
6,01332044
edited Dec 23 '18 at 17:52
answered Dec 23 '18 at 17:36
StatsStudentStatsStudent
6,01332044
answered Dec 23 '18 at 17:36
StatsStudentStatsStudent
6,01332044
answered Dec 23 '18 at 17:36
StatsStudentStatsStudent
6,01332044
6,01332044
$begingroup$
Are you using "To be sure, [...]" to mean "You can be sure of this because [...]"? (I ask because "To be sure, [...]" means something quite different in English, which makes your usage very confusing. If you didn't intend the idiomatic meaning, I recommend finding a different phrase to express what you meant.)
$endgroup$
– ruakh
Dec 24 '18 at 4:07
$begingroup$
@ruakh, indeed the intention was idiomatic. It is correctly used to mean "certainly, undoubtedly, admittedly." I will keep it.
$endgroup$
– StatsStudent
Dec 26 '18 at 17:27
$begingroup$
But "admittedly" makes no sense in this context. You must be misunderstanding the idiom. :-/
$endgroup$
– ruakh
Dec 26 '18 at 18:02
$begingroup$
No, I have a perfect understanding. I'm using the idiomatic meaning of "certainly" or "undoubtedly" - not "admittedly."
$endgroup$
– StatsStudent
Dec 26 '18 at 18:03
$begingroup$
That's not a separate meaning. You can use "certainly" to mean "to be sure" (for example, "To be sure, not everyone does it; but most people do" can be rephrased as "Certainly, not everyone does it; but most people do"); but you can't use "to be sure" in all cases where you can use "certainly" (for example, "It's certainly possible; would you like us to do it now?" can't be rephrased as "To be sure, it's possible; would you like us to do it now?"). If the word "admittedly" doesn't work, then the phrase "to be sure" doesn't work, either.
$endgroup$
– ruakh
Dec 26 '18 at 18:38
|
show 2 more comments
$begingroup$
Are you using "To be sure, [...]" to mean "You can be sure of this because [...]"? (I ask because "To be sure, [...]" means something quite different in English, which makes your usage very confusing. If you didn't intend the idiomatic meaning, I recommend finding a different phrase to express what you meant.)
$endgroup$
– ruakh
Dec 24 '18 at 4:07
$begingroup$
@ruakh, indeed the intention was idiomatic. It is correctly used to mean "certainly, undoubtedly, admittedly." I will keep it.
$endgroup$
– StatsStudent
Dec 26 '18 at 17:27
$begingroup$
But "admittedly" makes no sense in this context. You must be misunderstanding the idiom. :-/
$endgroup$
– ruakh
Dec 26 '18 at 18:02
$begingroup$
No, I have a perfect understanding. I'm using the idiomatic meaning of "certainly" or "undoubtedly" - not "admittedly."
$endgroup$
– StatsStudent
Dec 26 '18 at 18:03
$begingroup$
That's not a separate meaning. You can use "certainly" to mean "to be sure" (for example, "To be sure, not everyone does it; but most people do" can be rephrased as "Certainly, not everyone does it; but most people do"); but you can't use "to be sure" in all cases where you can use "certainly" (for example, "It's certainly possible; would you like us to do it now?" can't be rephrased as "To be sure, it's possible; would you like us to do it now?"). If the word "admittedly" doesn't work, then the phrase "to be sure" doesn't work, either.
$endgroup$
– ruakh
Dec 26 '18 at 18:38
$begingroup$
Are you using "To be sure, [...]" to mean "You can be sure of this because [...]"? (I ask because "To be sure, [...]" means something quite different in English, which makes your usage very confusing. If you didn't intend the idiomatic meaning, I recommend finding a different phrase to express what you meant.)
$endgroup$
– ruakh
Dec 24 '18 at 4:07
$begingroup$
Are you using "To be sure, [...]" to mean "You can be sure of this because [...]"? (I ask because "To be sure, [...]" means something quite different in English, which makes your usage very confusing. If you didn't intend the idiomatic meaning, I recommend finding a different phrase to express what you meant.)
$endgroup$
– ruakh
Dec 24 '18 at 4:07
$begingroup$
@ruakh, indeed the intention was idiomatic. It is correctly used to mean "certainly, undoubtedly, admittedly." I will keep it.
$endgroup$
– StatsStudent
Dec 26 '18 at 17:27
$begingroup$
@ruakh, indeed the intention was idiomatic. It is correctly used to mean "certainly, undoubtedly, admittedly." I will keep it.
$endgroup$
– StatsStudent
Dec 26 '18 at 17:27
$begingroup$
But "admittedly" makes no sense in this context. You must be misunderstanding the idiom. :-/
$endgroup$
– ruakh
Dec 26 '18 at 18:02
$begingroup$
But "admittedly" makes no sense in this context. You must be misunderstanding the idiom. :-/
$endgroup$
– ruakh
Dec 26 '18 at 18:02
$begingroup$
No, I have a perfect understanding. I'm using the idiomatic meaning of "certainly" or "undoubtedly" - not "admittedly."
$endgroup$
– StatsStudent
Dec 26 '18 at 18:03
$begingroup$
No, I have a perfect understanding. I'm using the idiomatic meaning of "certainly" or "undoubtedly" - not "admittedly."
$endgroup$
– StatsStudent
Dec 26 '18 at 18:03
$begingroup$
That's not a separate meaning. You can use "certainly" to mean "to be sure" (for example, "To be sure, not everyone does it; but most people do" can be rephrased as "Certainly, not everyone does it; but most people do"); but you can't use "to be sure" in all cases where you can use "certainly" (for example, "It's certainly possible; would you like us to do it now?" can't be rephrased as "To be sure, it's possible; would you like us to do it now?"). If the word "admittedly" doesn't work, then the phrase "to be sure" doesn't work, either.
$endgroup$
– ruakh
Dec 26 '18 at 18:38
$begingroup$
That's not a separate meaning. You can use "certainly" to mean "to be sure" (for example, "To be sure, not everyone does it; but most people do" can be rephrased as "Certainly, not everyone does it; but most people do"); but you can't use "to be sure" in all cases where you can use "certainly" (for example, "It's certainly possible; would you like us to do it now?" can't be rephrased as "To be sure, it's possible; would you like us to do it now?"). If the word "admittedly" doesn't work, then the phrase "to be sure" doesn't work, either.
$endgroup$
– ruakh
Dec 26 '18 at 18:38
|
show 2 more comments
$begingroup$
The PMF in the question, $f(x_i)=p^{x_i}(1-p)^{1-x_i}$ belongs to Bernoulli distribution, where $x_i$ is a binary variable. Yours is the PMF of Binomial distribution, and $x_i$'s are Binomial RVs ($n$ of them actually) with parameters $(m,p)$.
answered Dec 23 '18 at 17:39
gunesgunes
5,7601115
$endgroup$
add a comment |
$begingroup$
The PMF in the question, $f(x_i)=p^{x_i}(1-p)^{1-x_i}$ belongs to Bernoulli distribution, where $x_i$ is a binary variable. Yours is the PMF of Binomial distribution, and $x_i$'s are Binomial RVs ($n$ of them actually) with parameters $(m,p)$.
answered Dec 23 '18 at 17:39
gunesgunes
5,7601115
$endgroup$
add a comment |
$begingroup$
The PMF in the question, $f(x_i)=p^{x_i}(1-p)^{1-x_i}$ belongs to Bernoulli distribution, where $x_i$ is a binary variable. Yours is the PMF of Binomial distribution, and $x_i$'s are Binomial RVs ($n$ of them actually) with parameters $(m,p)$.
answered Dec 23 '18 at 17:39
gunesgunes
5,7601115
$endgroup$
The PMF in the question, $f(x_i)=p^{x_i}(1-p)^{1-x_i}$ belongs to Bernoulli distribution, where $x_i$ is a binary variable. Yours is the PMF of Binomial distribution, and $x_i$'s are Binomial RVs ($n$ of them actually) with parameters $(m,p)$.
answered Dec 23 '18 at 17:39
gunesgunes
5,7601115
answered Dec 23 '18 at 17:39
gunesgunes
5,7601115
answered Dec 23 '18 at 17:39
gunesgunes
5,7601115
answered Dec 23 '18 at 17:39
gunesgunes
5,7601115
5,7601115
add a comment |
add a comment |
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