Finding mode in Binomial distribution












6












$begingroup$


Suppose that $X$ has the Binomial distribution with parameters $n,p$ . How can I show that if $(n+1)p$ is integer then
$X$ has two mode that is $(n+1)p$ or $(n+1)p-1?$






probability statistics binomial-distribution







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: compute the ratio $b(n,p;k+1)/b(n,p;k)$ and check that this ratio is $gt1$ for every $klt k^*$ and $leqslant1$ for every $kgeqslant k^*$, for some integer $k^*$.
    $endgroup$
    – Did
    Mar 8 '12 at 16:27
















6












$begingroup$


Suppose that $X$ has the Binomial distribution with parameters $n,p$ . How can I show that if $(n+1)p$ is integer then
$X$ has two mode that is $(n+1)p$ or $(n+1)p-1?$






probability statistics binomial-distribution







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: compute the ratio $b(n,p;k+1)/b(n,p;k)$ and check that this ratio is $gt1$ for every $klt k^*$ and $leqslant1$ for every $kgeqslant k^*$, for some integer $k^*$.
    $endgroup$
    – Did
    Mar 8 '12 at 16:27














6












6








6


4



$begingroup$


Suppose that $X$ has the Binomial distribution with parameters $n,p$ . How can I show that if $(n+1)p$ is integer then
$X$ has two mode that is $(n+1)p$ or $(n+1)p-1?$






probability statistics binomial-distribution







share|cite|improve this question











$endgroup$




Suppose that $X$ has the Binomial distribution with parameters $n,p$ . How can I show that if $(n+1)p$ is integer then
$X$ has two mode that is $(n+1)p$ or $(n+1)p-1?$






probability statistics binomial-distribution




probability statistics binomial-distribution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 2 '17 at 10:08









Leila

3,48553056




3,48553056










asked Mar 8 '12 at 16:03









hadisanjihadisanji

5121511




5121511












  • $begingroup$
    Hint: compute the ratio $b(n,p;k+1)/b(n,p;k)$ and check that this ratio is $gt1$ for every $klt k^*$ and $leqslant1$ for every $kgeqslant k^*$, for some integer $k^*$.
    $endgroup$
    – Did
    Mar 8 '12 at 16:27


















  • $begingroup$
    Hint: compute the ratio $b(n,p;k+1)/b(n,p;k)$ and check that this ratio is $gt1$ for every $klt k^*$ and $leqslant1$ for every $kgeqslant k^*$, for some integer $k^*$.
    $endgroup$
    – Did
    Mar 8 '12 at 16:27
















$begingroup$
Hint: compute the ratio $b(n,p;k+1)/b(n,p;k)$ and check that this ratio is $gt1$ for every $klt k^*$ and $leqslant1$ for every $kgeqslant k^*$, for some integer $k^*$.
$endgroup$
– Did
Mar 8 '12 at 16:27




$begingroup$
Hint: compute the ratio $b(n,p;k+1)/b(n,p;k)$ and check that this ratio is $gt1$ for every $klt k^*$ and $leqslant1$ for every $kgeqslant k^*$, for some integer $k^*$.
$endgroup$
– Did
Mar 8 '12 at 16:27










1 Answer
1






active

oldest

votes


















9












$begingroup$

Let $a_k=P(X=k)$, we have
$$a_k=binom{n}{k}p^kq^{n-k}qquadtext{and}qquad a_{k+1}=binom{n}{k+1}p^{k+1}q^{n-k-
1},$$
where as usual $q=1-p$ in binomial distribution.



We calculate the ratio $dfrac{a_{k+1}}{a_k}$. Note that
$frac{binom{n}{k+1}}{binom{n}{k}}$
simplifies to
$frac{n-k}{k+1},$
and therefore
$$frac{a_{k+1}}{a_k}=frac{n-k}{k+1}cdotfrac{p}{q}=frac{n-k}{k+1}cdotfrac{p}{1-p}.$$



From this equation we can follow:



$$begin{align}
k > (n+1)p-1 implies a_{k+1} < a_k \
k = (n+1)p-1 implies a_{k+1} = a_k \
k < (n+1)p-1 implies a_{k+1} > a_k
end{align}$$



The calculation (almost) says that we have equality of two consecutive probabilities precisely if $a_{k+1}=a_k$, that is, if $k=np+p-1$. Note that $k=np+p-1$ implies that $np+p-1$ is an integer.



So if $k=np+p-1$ is not an integer, there is a single mode; and if $k=np+p-1$ is an integer, there are two modes, at $np+p-1$ and at $np+p$.



Not quite! We have been a little casual in our algebra. We have not paid attention to whether we might be multiplying or dividing by $0$. We also have casually accepted what the algebra seems to say, without doing a reality check.



Suppose that $p=0$. Then $np+p-1$ is an integer, namely $-1$. But whatever $n$ is, there is a single mode, namely $k=0$. In all other situations where $np+p-1$ is an integer, the $k$ we have identified is non-negative.



However, suppose that $p=1$. Again, $np+p-1$ is an integer, and again there is no double mode. The largest $a_k$ occurs at one place only, namely $k=n$, since $np+p$ is in this case beyond our range.



That completes the analysis when $np+p-1$ is an integer. When it is not, the analysis is simple. There is a single mode, at $lfloor np+prfloor$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I am solving a similar exercise and I have some doubts: why taking the ratio $frac{p_X(k+1)}{p_X(k)}$ gives you the mode (most probable value), which is defined as $sup_{x in R_X} p_X(x)$?
    $endgroup$
    – user16924
    Sep 7 '14 at 23:26








  • 2




    $begingroup$
    The sup is in this case a max, since the random variable takes on integer values. Apart from a couple of "degenerate" cases pointed out in the answer, the probabilities rise and then fall. The ratio of consecutive terms is therefore $gt 1$ for a while, then $lt 1$, except that in somewhat unusual cases we can have ratio $1$, so two consecutive values each qualify as a mode. Looking at the ratios tells us when the probability has reached a maximum.
    $endgroup$
    – André Nicolas
    Sep 8 '14 at 1:38










  • $begingroup$
    Thanks very much for the explanation.
    $endgroup$
    – user16924
    Sep 8 '14 at 2:47










  • $begingroup$
    You are welcome. That was an overview. The detail is in the answer above.
    $endgroup$
    – André Nicolas
    Sep 8 '14 at 2:48











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1 Answer
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9












$begingroup$

Let $a_k=P(X=k)$, we have
$$a_k=binom{n}{k}p^kq^{n-k}qquadtext{and}qquad a_{k+1}=binom{n}{k+1}p^{k+1}q^{n-k-
1},$$
where as usual $q=1-p$ in binomial distribution.



We calculate the ratio $dfrac{a_{k+1}}{a_k}$. Note that
$frac{binom{n}{k+1}}{binom{n}{k}}$
simplifies to
$frac{n-k}{k+1},$
and therefore
$$frac{a_{k+1}}{a_k}=frac{n-k}{k+1}cdotfrac{p}{q}=frac{n-k}{k+1}cdotfrac{p}{1-p}.$$



From this equation we can follow:



$$begin{align}
k > (n+1)p-1 implies a_{k+1} < a_k \
k = (n+1)p-1 implies a_{k+1} = a_k \
k < (n+1)p-1 implies a_{k+1} > a_k
end{align}$$



The calculation (almost) says that we have equality of two consecutive probabilities precisely if $a_{k+1}=a_k$, that is, if $k=np+p-1$. Note that $k=np+p-1$ implies that $np+p-1$ is an integer.



So if $k=np+p-1$ is not an integer, there is a single mode; and if $k=np+p-1$ is an integer, there are two modes, at $np+p-1$ and at $np+p$.



Not quite! We have been a little casual in our algebra. We have not paid attention to whether we might be multiplying or dividing by $0$. We also have casually accepted what the algebra seems to say, without doing a reality check.



Suppose that $p=0$. Then $np+p-1$ is an integer, namely $-1$. But whatever $n$ is, there is a single mode, namely $k=0$. In all other situations where $np+p-1$ is an integer, the $k$ we have identified is non-negative.



However, suppose that $p=1$. Again, $np+p-1$ is an integer, and again there is no double mode. The largest $a_k$ occurs at one place only, namely $k=n$, since $np+p$ is in this case beyond our range.



That completes the analysis when $np+p-1$ is an integer. When it is not, the analysis is simple. There is a single mode, at $lfloor np+prfloor$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I am solving a similar exercise and I have some doubts: why taking the ratio $frac{p_X(k+1)}{p_X(k)}$ gives you the mode (most probable value), which is defined as $sup_{x in R_X} p_X(x)$?
    $endgroup$
    – user16924
    Sep 7 '14 at 23:26








  • 2




    $begingroup$
    The sup is in this case a max, since the random variable takes on integer values. Apart from a couple of "degenerate" cases pointed out in the answer, the probabilities rise and then fall. The ratio of consecutive terms is therefore $gt 1$ for a while, then $lt 1$, except that in somewhat unusual cases we can have ratio $1$, so two consecutive values each qualify as a mode. Looking at the ratios tells us when the probability has reached a maximum.
    $endgroup$
    – André Nicolas
    Sep 8 '14 at 1:38










  • $begingroup$
    Thanks very much for the explanation.
    $endgroup$
    – user16924
    Sep 8 '14 at 2:47










  • $begingroup$
    You are welcome. That was an overview. The detail is in the answer above.
    $endgroup$
    – André Nicolas
    Sep 8 '14 at 2:48
















9












$begingroup$

Let $a_k=P(X=k)$, we have
$$a_k=binom{n}{k}p^kq^{n-k}qquadtext{and}qquad a_{k+1}=binom{n}{k+1}p^{k+1}q^{n-k-
1},$$
where as usual $q=1-p$ in binomial distribution.



We calculate the ratio $dfrac{a_{k+1}}{a_k}$. Note that
$frac{binom{n}{k+1}}{binom{n}{k}}$
simplifies to
$frac{n-k}{k+1},$
and therefore
$$frac{a_{k+1}}{a_k}=frac{n-k}{k+1}cdotfrac{p}{q}=frac{n-k}{k+1}cdotfrac{p}{1-p}.$$



From this equation we can follow:



$$begin{align}
k > (n+1)p-1 implies a_{k+1} < a_k \
k = (n+1)p-1 implies a_{k+1} = a_k \
k < (n+1)p-1 implies a_{k+1} > a_k
end{align}$$



The calculation (almost) says that we have equality of two consecutive probabilities precisely if $a_{k+1}=a_k$, that is, if $k=np+p-1$. Note that $k=np+p-1$ implies that $np+p-1$ is an integer.



So if $k=np+p-1$ is not an integer, there is a single mode; and if $k=np+p-1$ is an integer, there are two modes, at $np+p-1$ and at $np+p$.



Not quite! We have been a little casual in our algebra. We have not paid attention to whether we might be multiplying or dividing by $0$. We also have casually accepted what the algebra seems to say, without doing a reality check.



Suppose that $p=0$. Then $np+p-1$ is an integer, namely $-1$. But whatever $n$ is, there is a single mode, namely $k=0$. In all other situations where $np+p-1$ is an integer, the $k$ we have identified is non-negative.



However, suppose that $p=1$. Again, $np+p-1$ is an integer, and again there is no double mode. The largest $a_k$ occurs at one place only, namely $k=n$, since $np+p$ is in this case beyond our range.



That completes the analysis when $np+p-1$ is an integer. When it is not, the analysis is simple. There is a single mode, at $lfloor np+prfloor$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I am solving a similar exercise and I have some doubts: why taking the ratio $frac{p_X(k+1)}{p_X(k)}$ gives you the mode (most probable value), which is defined as $sup_{x in R_X} p_X(x)$?
    $endgroup$
    – user16924
    Sep 7 '14 at 23:26








  • 2




    $begingroup$
    The sup is in this case a max, since the random variable takes on integer values. Apart from a couple of "degenerate" cases pointed out in the answer, the probabilities rise and then fall. The ratio of consecutive terms is therefore $gt 1$ for a while, then $lt 1$, except that in somewhat unusual cases we can have ratio $1$, so two consecutive values each qualify as a mode. Looking at the ratios tells us when the probability has reached a maximum.
    $endgroup$
    – André Nicolas
    Sep 8 '14 at 1:38










  • $begingroup$
    Thanks very much for the explanation.
    $endgroup$
    – user16924
    Sep 8 '14 at 2:47










  • $begingroup$
    You are welcome. That was an overview. The detail is in the answer above.
    $endgroup$
    – André Nicolas
    Sep 8 '14 at 2:48














9












9








9





$begingroup$

Let $a_k=P(X=k)$, we have
$$a_k=binom{n}{k}p^kq^{n-k}qquadtext{and}qquad a_{k+1}=binom{n}{k+1}p^{k+1}q^{n-k-
1},$$
where as usual $q=1-p$ in binomial distribution.



We calculate the ratio $dfrac{a_{k+1}}{a_k}$. Note that
$frac{binom{n}{k+1}}{binom{n}{k}}$
simplifies to
$frac{n-k}{k+1},$
and therefore
$$frac{a_{k+1}}{a_k}=frac{n-k}{k+1}cdotfrac{p}{q}=frac{n-k}{k+1}cdotfrac{p}{1-p}.$$



From this equation we can follow:



$$begin{align}
k > (n+1)p-1 implies a_{k+1} < a_k \
k = (n+1)p-1 implies a_{k+1} = a_k \
k < (n+1)p-1 implies a_{k+1} > a_k
end{align}$$



The calculation (almost) says that we have equality of two consecutive probabilities precisely if $a_{k+1}=a_k$, that is, if $k=np+p-1$. Note that $k=np+p-1$ implies that $np+p-1$ is an integer.



So if $k=np+p-1$ is not an integer, there is a single mode; and if $k=np+p-1$ is an integer, there are two modes, at $np+p-1$ and at $np+p$.



Not quite! We have been a little casual in our algebra. We have not paid attention to whether we might be multiplying or dividing by $0$. We also have casually accepted what the algebra seems to say, without doing a reality check.



Suppose that $p=0$. Then $np+p-1$ is an integer, namely $-1$. But whatever $n$ is, there is a single mode, namely $k=0$. In all other situations where $np+p-1$ is an integer, the $k$ we have identified is non-negative.



However, suppose that $p=1$. Again, $np+p-1$ is an integer, and again there is no double mode. The largest $a_k$ occurs at one place only, namely $k=n$, since $np+p$ is in this case beyond our range.



That completes the analysis when $np+p-1$ is an integer. When it is not, the analysis is simple. There is a single mode, at $lfloor np+prfloor$.






share|cite|improve this answer











$endgroup$



Let $a_k=P(X=k)$, we have
$$a_k=binom{n}{k}p^kq^{n-k}qquadtext{and}qquad a_{k+1}=binom{n}{k+1}p^{k+1}q^{n-k-
1},$$
where as usual $q=1-p$ in binomial distribution.



We calculate the ratio $dfrac{a_{k+1}}{a_k}$. Note that
$frac{binom{n}{k+1}}{binom{n}{k}}$
simplifies to
$frac{n-k}{k+1},$
and therefore
$$frac{a_{k+1}}{a_k}=frac{n-k}{k+1}cdotfrac{p}{q}=frac{n-k}{k+1}cdotfrac{p}{1-p}.$$



From this equation we can follow:



$$begin{align}
k > (n+1)p-1 implies a_{k+1} < a_k \
k = (n+1)p-1 implies a_{k+1} = a_k \
k < (n+1)p-1 implies a_{k+1} > a_k
end{align}$$



The calculation (almost) says that we have equality of two consecutive probabilities precisely if $a_{k+1}=a_k$, that is, if $k=np+p-1$. Note that $k=np+p-1$ implies that $np+p-1$ is an integer.



So if $k=np+p-1$ is not an integer, there is a single mode; and if $k=np+p-1$ is an integer, there are two modes, at $np+p-1$ and at $np+p$.



Not quite! We have been a little casual in our algebra. We have not paid attention to whether we might be multiplying or dividing by $0$. We also have casually accepted what the algebra seems to say, without doing a reality check.



Suppose that $p=0$. Then $np+p-1$ is an integer, namely $-1$. But whatever $n$ is, there is a single mode, namely $k=0$. In all other situations where $np+p-1$ is an integer, the $k$ we have identified is non-negative.



However, suppose that $p=1$. Again, $np+p-1$ is an integer, and again there is no double mode. The largest $a_k$ occurs at one place only, namely $k=n$, since $np+p$ is in this case beyond our range.



That completes the analysis when $np+p-1$ is an integer. When it is not, the analysis is simple. There is a single mode, at $lfloor np+prfloor$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 7 at 6:10









rekcah

33




33










answered Mar 8 '12 at 16:44









André NicolasAndré Nicolas

452k36423808




452k36423808












  • $begingroup$
    I am solving a similar exercise and I have some doubts: why taking the ratio $frac{p_X(k+1)}{p_X(k)}$ gives you the mode (most probable value), which is defined as $sup_{x in R_X} p_X(x)$?
    $endgroup$
    – user16924
    Sep 7 '14 at 23:26








  • 2




    $begingroup$
    The sup is in this case a max, since the random variable takes on integer values. Apart from a couple of "degenerate" cases pointed out in the answer, the probabilities rise and then fall. The ratio of consecutive terms is therefore $gt 1$ for a while, then $lt 1$, except that in somewhat unusual cases we can have ratio $1$, so two consecutive values each qualify as a mode. Looking at the ratios tells us when the probability has reached a maximum.
    $endgroup$
    – André Nicolas
    Sep 8 '14 at 1:38










  • $begingroup$
    Thanks very much for the explanation.
    $endgroup$
    – user16924
    Sep 8 '14 at 2:47










  • $begingroup$
    You are welcome. That was an overview. The detail is in the answer above.
    $endgroup$
    – André Nicolas
    Sep 8 '14 at 2:48


















  • $begingroup$
    I am solving a similar exercise and I have some doubts: why taking the ratio $frac{p_X(k+1)}{p_X(k)}$ gives you the mode (most probable value), which is defined as $sup_{x in R_X} p_X(x)$?
    $endgroup$
    – user16924
    Sep 7 '14 at 23:26








  • 2




    $begingroup$
    The sup is in this case a max, since the random variable takes on integer values. Apart from a couple of "degenerate" cases pointed out in the answer, the probabilities rise and then fall. The ratio of consecutive terms is therefore $gt 1$ for a while, then $lt 1$, except that in somewhat unusual cases we can have ratio $1$, so two consecutive values each qualify as a mode. Looking at the ratios tells us when the probability has reached a maximum.
    $endgroup$
    – André Nicolas
    Sep 8 '14 at 1:38










  • $begingroup$
    Thanks very much for the explanation.
    $endgroup$
    – user16924
    Sep 8 '14 at 2:47










  • $begingroup$
    You are welcome. That was an overview. The detail is in the answer above.
    $endgroup$
    – André Nicolas
    Sep 8 '14 at 2:48
















$begingroup$
I am solving a similar exercise and I have some doubts: why taking the ratio $frac{p_X(k+1)}{p_X(k)}$ gives you the mode (most probable value), which is defined as $sup_{x in R_X} p_X(x)$?
$endgroup$
– user16924
Sep 7 '14 at 23:26






$begingroup$
I am solving a similar exercise and I have some doubts: why taking the ratio $frac{p_X(k+1)}{p_X(k)}$ gives you the mode (most probable value), which is defined as $sup_{x in R_X} p_X(x)$?
$endgroup$
– user16924
Sep 7 '14 at 23:26






2




2




$begingroup$
The sup is in this case a max, since the random variable takes on integer values. Apart from a couple of "degenerate" cases pointed out in the answer, the probabilities rise and then fall. The ratio of consecutive terms is therefore $gt 1$ for a while, then $lt 1$, except that in somewhat unusual cases we can have ratio $1$, so two consecutive values each qualify as a mode. Looking at the ratios tells us when the probability has reached a maximum.
$endgroup$
– André Nicolas
Sep 8 '14 at 1:38




$begingroup$
The sup is in this case a max, since the random variable takes on integer values. Apart from a couple of "degenerate" cases pointed out in the answer, the probabilities rise and then fall. The ratio of consecutive terms is therefore $gt 1$ for a while, then $lt 1$, except that in somewhat unusual cases we can have ratio $1$, so two consecutive values each qualify as a mode. Looking at the ratios tells us when the probability has reached a maximum.
$endgroup$
– André Nicolas
Sep 8 '14 at 1:38












$begingroup$
Thanks very much for the explanation.
$endgroup$
– user16924
Sep 8 '14 at 2:47




$begingroup$
Thanks very much for the explanation.
$endgroup$
– user16924
Sep 8 '14 at 2:47












$begingroup$
You are welcome. That was an overview. The detail is in the answer above.
$endgroup$
– André Nicolas
Sep 8 '14 at 2:48




$begingroup$
You are welcome. That was an overview. The detail is in the answer above.
$endgroup$
– André Nicolas
Sep 8 '14 at 2:48


















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Management

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Management From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Manager" redirects here. For other uses, see Management (disambiguation) and Manager (disambiguation). An organization chart for the United States Coast Guard shows the hierarchy of managerial roles in that organization. Business administration Management of a business Accounting Management accounting Financial accounting Financial audit Business entities Cooperative Corporation Limited liability company Partnership Sole proprietorship State-owned enterprise Corporate governance Annual general meeting Board of directors Supervisory board Advisory board Audit committee Corporate law Commercial law Constitutional documents Contract Corporate crime Corporate liability Insolvency law International trade law Mergers and acquisitions Economics Commodity Pu
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Investment

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Investment From Wikipedia, the free encyclopedia Jump to navigation Jump to search This article is about investment in finance. For investment in macroeconomics, see Investment (macroeconomics). For other uses, see Investment (disambiguation). "Invest" redirects here. For the term in meteorology, see Invest (meteorology). This article needs additional citations for verification . Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed. ( February 2017 ) (Learn how and when to remove this template message) In general, to invest is to distribute money in the expectation of some benefit in the future – for example, investment in durable goods, in real estate by the service industry, in factories for manufacturing, in product development, and in research and development. However, this article focuses specifically on investment in financial
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