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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active

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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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1300-talet

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Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
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1300-talet

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Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
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Display a custom attribute below product name in the front-end Magento 1.9.3.8

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0 I need to display a custom attribute bellow the product name and I don't know how can I call this attribute. I am editing the file /app/design/frontend/templatemela/MAG100219_1/template/catalog/product/view.phtml But how can I call this attribute in php? I tried calling a string echo with the name I put in the attribute, but it didn't work. Code is as below: <div class="page-title product-title"> <h1><?php echo $_helper->productAttribute($_product, $_product->getName(), 'name') ?></h1> </div> <div class="codigoinmetro"> <h2><?php echo $_helper->productAttribute($_product, $_product->getName(), 'registroinmetro') ?></h2> </div> php attrib...
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