Showing monotonicity for ratio of binomial pmf and tail cdf












2












$begingroup$


I'm interested in showing for $Xsimtext{Bin}(n,p)$, $pin(0,1)$ that when $xgeq np$,
$$
frac{P(X=x)}{P(Xgeq x)}leq frac{P(X=x+1)}{P(Xgeq x+1)}
$$
I've verified using numerical simulations, but can't seem to get the algebra to work out.



Below I plot the ratio $P(X=x)/P(Xgeq x)$ for values of $xgeq lfloor nprfloor$. We can see that the ratio is increasing with $x$. Furthermore, I plot the difference between the ratio evaluated at $x+1$ (the right hand side above) and the ratio evaluated at $x$ (the left hand side above) for varying $x$. The difference is also positive, though the difference increases for $x$ close to $np$ and then decreases thereafter to zero.



Numerical Simulations



n=500

p=0.3

x=seq(floor(n*p),n)

ratio=dbinom(x,n,p)/pbinom(x-1,n,p,lower.tail = FALSE)

diff=dbinom(x,n,p)/pbinom(x-1,n,p,lower.tail = FALSE)-dbinom(x-1,n,p)/pbinom(x-2,n,p,lower.tail = FALSE)



plot(x,ratio)

plot(x,diff)









inequality probability-distributions binomial-distribution monotone-functions







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$endgroup$












  • $begingroup$
    A good place to start would be writing the actual values for $P(X=x),P(X=x+1),P(Xgeq x),P(Xgeq x+1)$ so you know what you're trying to prove. Can you do this?
    $endgroup$
    – pwerth
    Jan 23 at 21:40










  • $begingroup$
    Yes, I've done that as well. Using just brute force algebra (splitting up components, etc.) didn't yield me much.
    $endgroup$
    – stats134711
    Jan 23 at 21:43
















2












$begingroup$


I'm interested in showing for $Xsimtext{Bin}(n,p)$, $pin(0,1)$ that when $xgeq np$,
$$
frac{P(X=x)}{P(Xgeq x)}leq frac{P(X=x+1)}{P(Xgeq x+1)}
$$
I've verified using numerical simulations, but can't seem to get the algebra to work out.



Below I plot the ratio $P(X=x)/P(Xgeq x)$ for values of $xgeq lfloor nprfloor$. We can see that the ratio is increasing with $x$. Furthermore, I plot the difference between the ratio evaluated at $x+1$ (the right hand side above) and the ratio evaluated at $x$ (the left hand side above) for varying $x$. The difference is also positive, though the difference increases for $x$ close to $np$ and then decreases thereafter to zero.



Numerical Simulations



n=500

p=0.3

x=seq(floor(n*p),n)

ratio=dbinom(x,n,p)/pbinom(x-1,n,p,lower.tail = FALSE)

diff=dbinom(x,n,p)/pbinom(x-1,n,p,lower.tail = FALSE)-dbinom(x-1,n,p)/pbinom(x-2,n,p,lower.tail = FALSE)



plot(x,ratio)

plot(x,diff)









inequality probability-distributions binomial-distribution monotone-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    A good place to start would be writing the actual values for $P(X=x),P(X=x+1),P(Xgeq x),P(Xgeq x+1)$ so you know what you're trying to prove. Can you do this?
    $endgroup$
    – pwerth
    Jan 23 at 21:40










  • $begingroup$
    Yes, I've done that as well. Using just brute force algebra (splitting up components, etc.) didn't yield me much.
    $endgroup$
    – stats134711
    Jan 23 at 21:43














2












2








2





$begingroup$


I'm interested in showing for $Xsimtext{Bin}(n,p)$, $pin(0,1)$ that when $xgeq np$,
$$
frac{P(X=x)}{P(Xgeq x)}leq frac{P(X=x+1)}{P(Xgeq x+1)}
$$
I've verified using numerical simulations, but can't seem to get the algebra to work out.



Below I plot the ratio $P(X=x)/P(Xgeq x)$ for values of $xgeq lfloor nprfloor$. We can see that the ratio is increasing with $x$. Furthermore, I plot the difference between the ratio evaluated at $x+1$ (the right hand side above) and the ratio evaluated at $x$ (the left hand side above) for varying $x$. The difference is also positive, though the difference increases for $x$ close to $np$ and then decreases thereafter to zero.



Numerical Simulations



n=500

p=0.3

x=seq(floor(n*p),n)

ratio=dbinom(x,n,p)/pbinom(x-1,n,p,lower.tail = FALSE)

diff=dbinom(x,n,p)/pbinom(x-1,n,p,lower.tail = FALSE)-dbinom(x-1,n,p)/pbinom(x-2,n,p,lower.tail = FALSE)



plot(x,ratio)

plot(x,diff)









inequality probability-distributions binomial-distribution monotone-functions







share|cite|improve this question











$endgroup$




I'm interested in showing for $Xsimtext{Bin}(n,p)$, $pin(0,1)$ that when $xgeq np$,
$$
frac{P(X=x)}{P(Xgeq x)}leq frac{P(X=x+1)}{P(Xgeq x+1)}
$$
I've verified using numerical simulations, but can't seem to get the algebra to work out.



Below I plot the ratio $P(X=x)/P(Xgeq x)$ for values of $xgeq lfloor nprfloor$. We can see that the ratio is increasing with $x$. Furthermore, I plot the difference between the ratio evaluated at $x+1$ (the right hand side above) and the ratio evaluated at $x$ (the left hand side above) for varying $x$. The difference is also positive, though the difference increases for $x$ close to $np$ and then decreases thereafter to zero.



Numerical Simulations



n=500

p=0.3

x=seq(floor(n*p),n)

ratio=dbinom(x,n,p)/pbinom(x-1,n,p,lower.tail = FALSE)

diff=dbinom(x,n,p)/pbinom(x-1,n,p,lower.tail = FALSE)-dbinom(x-1,n,p)/pbinom(x-2,n,p,lower.tail = FALSE)



plot(x,ratio)

plot(x,diff)









inequality probability-distributions binomial-distribution monotone-functions




inequality probability-distributions binomial-distribution monotone-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 25 at 2:51







stats134711

















asked Jan 23 at 21:34









stats134711stats134711

187215




187215












  • $begingroup$
    A good place to start would be writing the actual values for $P(X=x),P(X=x+1),P(Xgeq x),P(Xgeq x+1)$ so you know what you're trying to prove. Can you do this?
    $endgroup$
    – pwerth
    Jan 23 at 21:40










  • $begingroup$
    Yes, I've done that as well. Using just brute force algebra (splitting up components, etc.) didn't yield me much.
    $endgroup$
    – stats134711
    Jan 23 at 21:43


















  • $begingroup$
    A good place to start would be writing the actual values for $P(X=x),P(X=x+1),P(Xgeq x),P(Xgeq x+1)$ so you know what you're trying to prove. Can you do this?
    $endgroup$
    – pwerth
    Jan 23 at 21:40










  • $begingroup$
    Yes, I've done that as well. Using just brute force algebra (splitting up components, etc.) didn't yield me much.
    $endgroup$
    – stats134711
    Jan 23 at 21:43
















$begingroup$
A good place to start would be writing the actual values for $P(X=x),P(X=x+1),P(Xgeq x),P(Xgeq x+1)$ so you know what you're trying to prove. Can you do this?
$endgroup$
– pwerth
Jan 23 at 21:40




$begingroup$
A good place to start would be writing the actual values for $P(X=x),P(X=x+1),P(Xgeq x),P(Xgeq x+1)$ so you know what you're trying to prove. Can you do this?
$endgroup$
– pwerth
Jan 23 at 21:40












$begingroup$
Yes, I've done that as well. Using just brute force algebra (splitting up components, etc.) didn't yield me much.
$endgroup$
– stats134711
Jan 23 at 21:43




$begingroup$
Yes, I've done that as well. Using just brute force algebra (splitting up components, etc.) didn't yield me much.
$endgroup$
– stats134711
Jan 23 at 21:43










1 Answer
1






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oldest

votes


















1












$begingroup$

We know that
$$
frac{P(X=x+1)}{P(X=x)}=frac{n-x}{x+1}frac{p}{1-p}
$$
Therefore, we will show that
$$
frac{P(Xgeq x+1)}{P(Xgeq x)}leq frac{n-x}{x+1}frac{p}{1-p}
$$
Multiplying the left hand side by $(1-p)/p$ and expanding out the numerator and denominator on the left hand side we have
begin{align}
frac{1-p}{p}frac{P(Xgeq x+1)}{P(Xgeq x)}&=frac{1-p}{p}frac{sum_{kgeq x+1}dbinom{n}{k}p^k(1-p)^{n-k}}{sum_{kgeq x}dbinom{n}{k}p^k(1-p)^{n-k}}\
&=frac{binom{n}{x+1}p^{x+1}(1-p)^{n-x-1}+binom{n}{x+2}p^{x+2}(1-p)^{n-x-2}+binom{n}{x+3}p^{x+3}(1-p)^{n-x-3}+cdots}{binom{n}{x}p^{x+1}(1-p)^{n-x-1}+binom{n}{x+1}p^{x+2}(1-p)^{n-x-2}+binom{n}{x+2}p^{x+3}(1-p)^{n-x-3}+cdots}tag{1}
end{align}



Note that the numerator has $n-x$ terms while the denominator has $n-x+1$ terms. Now we examine ratios of successive binomial coefficients. Note that
$$
frac{binom{n}{x+1}}{binom{n}{x}}=frac{n-x}{x+1}
$$
and for $kgeq 1$
$$
frac{binom{n}{x+1+k}}{binom{n}{x+k}}=frac{n-x-k}{x+1+k}leq frac{n-x}{x+1}
$$
Combining the last two results, we may write for $kgeq 0$
$$
binom{n}{x+1+k}leq frac{n-x}{x+1}binom{n}{x+k}
$$
Substituting this inequality for each binomial coefficient in the numerator of (1), we have
begin{align}
frac{1-p}{p}frac{P(Xgeq x+1)}{P(Xgeq x)}&=frac{binom{n}{x+1}p^{x+1}(1-p)^{n-x-1}+binom{n}{x+2}p^{x+2}(1-p)^{n-x-2}+binom{n}{x+3}p^{x+3}(1-p)^{n-x-3}+cdots}{binom{n}{x}p^{x+1}(1-p)^{n-x-1}+binom{n}{x+1}p^{x+2}(1-p)^{n-x-2}+binom{n}{x+2}p^{x+3}(1-p)^{n-x-3}+cdots}\
&leq frac{n-x}{x+1}frac{sum_{k=x+1}^n binom{n}{k-1}p^k(1-p)^{n-k} }{left[sum_{k=x+1}^n binom{n}{k-1}p^k(1-p)^{n-k}right] +p^{n+1}(1-p)^{-1} }\
&leq frac{n-x}{x+1}
end{align}
as desired.






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    1 Answer
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    1 Answer
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    active

    oldest

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    1












    $begingroup$

    We know that
    $$
    frac{P(X=x+1)}{P(X=x)}=frac{n-x}{x+1}frac{p}{1-p}
    $$
    Therefore, we will show that
    $$
    frac{P(Xgeq x+1)}{P(Xgeq x)}leq frac{n-x}{x+1}frac{p}{1-p}
    $$
    Multiplying the left hand side by $(1-p)/p$ and expanding out the numerator and denominator on the left hand side we have
    begin{align}
    frac{1-p}{p}frac{P(Xgeq x+1)}{P(Xgeq x)}&=frac{1-p}{p}frac{sum_{kgeq x+1}dbinom{n}{k}p^k(1-p)^{n-k}}{sum_{kgeq x}dbinom{n}{k}p^k(1-p)^{n-k}}\
    &=frac{binom{n}{x+1}p^{x+1}(1-p)^{n-x-1}+binom{n}{x+2}p^{x+2}(1-p)^{n-x-2}+binom{n}{x+3}p^{x+3}(1-p)^{n-x-3}+cdots}{binom{n}{x}p^{x+1}(1-p)^{n-x-1}+binom{n}{x+1}p^{x+2}(1-p)^{n-x-2}+binom{n}{x+2}p^{x+3}(1-p)^{n-x-3}+cdots}tag{1}
    end{align}



    Note that the numerator has $n-x$ terms while the denominator has $n-x+1$ terms. Now we examine ratios of successive binomial coefficients. Note that
    $$
    frac{binom{n}{x+1}}{binom{n}{x}}=frac{n-x}{x+1}
    $$
    and for $kgeq 1$
    $$
    frac{binom{n}{x+1+k}}{binom{n}{x+k}}=frac{n-x-k}{x+1+k}leq frac{n-x}{x+1}
    $$
    Combining the last two results, we may write for $kgeq 0$
    $$
    binom{n}{x+1+k}leq frac{n-x}{x+1}binom{n}{x+k}
    $$
    Substituting this inequality for each binomial coefficient in the numerator of (1), we have
    begin{align}
    frac{1-p}{p}frac{P(Xgeq x+1)}{P(Xgeq x)}&=frac{binom{n}{x+1}p^{x+1}(1-p)^{n-x-1}+binom{n}{x+2}p^{x+2}(1-p)^{n-x-2}+binom{n}{x+3}p^{x+3}(1-p)^{n-x-3}+cdots}{binom{n}{x}p^{x+1}(1-p)^{n-x-1}+binom{n}{x+1}p^{x+2}(1-p)^{n-x-2}+binom{n}{x+2}p^{x+3}(1-p)^{n-x-3}+cdots}\
    &leq frac{n-x}{x+1}frac{sum_{k=x+1}^n binom{n}{k-1}p^k(1-p)^{n-k} }{left[sum_{k=x+1}^n binom{n}{k-1}p^k(1-p)^{n-k}right] +p^{n+1}(1-p)^{-1} }\
    &leq frac{n-x}{x+1}
    end{align}
    as desired.






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    $endgroup$


















      1












      $begingroup$

      We know that
      $$
      frac{P(X=x+1)}{P(X=x)}=frac{n-x}{x+1}frac{p}{1-p}
      $$
      Therefore, we will show that
      $$
      frac{P(Xgeq x+1)}{P(Xgeq x)}leq frac{n-x}{x+1}frac{p}{1-p}
      $$
      Multiplying the left hand side by $(1-p)/p$ and expanding out the numerator and denominator on the left hand side we have
      begin{align}
      frac{1-p}{p}frac{P(Xgeq x+1)}{P(Xgeq x)}&=frac{1-p}{p}frac{sum_{kgeq x+1}dbinom{n}{k}p^k(1-p)^{n-k}}{sum_{kgeq x}dbinom{n}{k}p^k(1-p)^{n-k}}\
      &=frac{binom{n}{x+1}p^{x+1}(1-p)^{n-x-1}+binom{n}{x+2}p^{x+2}(1-p)^{n-x-2}+binom{n}{x+3}p^{x+3}(1-p)^{n-x-3}+cdots}{binom{n}{x}p^{x+1}(1-p)^{n-x-1}+binom{n}{x+1}p^{x+2}(1-p)^{n-x-2}+binom{n}{x+2}p^{x+3}(1-p)^{n-x-3}+cdots}tag{1}
      end{align}



      Note that the numerator has $n-x$ terms while the denominator has $n-x+1$ terms. Now we examine ratios of successive binomial coefficients. Note that
      $$
      frac{binom{n}{x+1}}{binom{n}{x}}=frac{n-x}{x+1}
      $$
      and for $kgeq 1$
      $$
      frac{binom{n}{x+1+k}}{binom{n}{x+k}}=frac{n-x-k}{x+1+k}leq frac{n-x}{x+1}
      $$
      Combining the last two results, we may write for $kgeq 0$
      $$
      binom{n}{x+1+k}leq frac{n-x}{x+1}binom{n}{x+k}
      $$
      Substituting this inequality for each binomial coefficient in the numerator of (1), we have
      begin{align}
      frac{1-p}{p}frac{P(Xgeq x+1)}{P(Xgeq x)}&=frac{binom{n}{x+1}p^{x+1}(1-p)^{n-x-1}+binom{n}{x+2}p^{x+2}(1-p)^{n-x-2}+binom{n}{x+3}p^{x+3}(1-p)^{n-x-3}+cdots}{binom{n}{x}p^{x+1}(1-p)^{n-x-1}+binom{n}{x+1}p^{x+2}(1-p)^{n-x-2}+binom{n}{x+2}p^{x+3}(1-p)^{n-x-3}+cdots}\
      &leq frac{n-x}{x+1}frac{sum_{k=x+1}^n binom{n}{k-1}p^k(1-p)^{n-k} }{left[sum_{k=x+1}^n binom{n}{k-1}p^k(1-p)^{n-k}right] +p^{n+1}(1-p)^{-1} }\
      &leq frac{n-x}{x+1}
      end{align}
      as desired.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        We know that
        $$
        frac{P(X=x+1)}{P(X=x)}=frac{n-x}{x+1}frac{p}{1-p}
        $$
        Therefore, we will show that
        $$
        frac{P(Xgeq x+1)}{P(Xgeq x)}leq frac{n-x}{x+1}frac{p}{1-p}
        $$
        Multiplying the left hand side by $(1-p)/p$ and expanding out the numerator and denominator on the left hand side we have
        begin{align}
        frac{1-p}{p}frac{P(Xgeq x+1)}{P(Xgeq x)}&=frac{1-p}{p}frac{sum_{kgeq x+1}dbinom{n}{k}p^k(1-p)^{n-k}}{sum_{kgeq x}dbinom{n}{k}p^k(1-p)^{n-k}}\
        &=frac{binom{n}{x+1}p^{x+1}(1-p)^{n-x-1}+binom{n}{x+2}p^{x+2}(1-p)^{n-x-2}+binom{n}{x+3}p^{x+3}(1-p)^{n-x-3}+cdots}{binom{n}{x}p^{x+1}(1-p)^{n-x-1}+binom{n}{x+1}p^{x+2}(1-p)^{n-x-2}+binom{n}{x+2}p^{x+3}(1-p)^{n-x-3}+cdots}tag{1}
        end{align}



        Note that the numerator has $n-x$ terms while the denominator has $n-x+1$ terms. Now we examine ratios of successive binomial coefficients. Note that
        $$
        frac{binom{n}{x+1}}{binom{n}{x}}=frac{n-x}{x+1}
        $$
        and for $kgeq 1$
        $$
        frac{binom{n}{x+1+k}}{binom{n}{x+k}}=frac{n-x-k}{x+1+k}leq frac{n-x}{x+1}
        $$
        Combining the last two results, we may write for $kgeq 0$
        $$
        binom{n}{x+1+k}leq frac{n-x}{x+1}binom{n}{x+k}
        $$
        Substituting this inequality for each binomial coefficient in the numerator of (1), we have
        begin{align}
        frac{1-p}{p}frac{P(Xgeq x+1)}{P(Xgeq x)}&=frac{binom{n}{x+1}p^{x+1}(1-p)^{n-x-1}+binom{n}{x+2}p^{x+2}(1-p)^{n-x-2}+binom{n}{x+3}p^{x+3}(1-p)^{n-x-3}+cdots}{binom{n}{x}p^{x+1}(1-p)^{n-x-1}+binom{n}{x+1}p^{x+2}(1-p)^{n-x-2}+binom{n}{x+2}p^{x+3}(1-p)^{n-x-3}+cdots}\
        &leq frac{n-x}{x+1}frac{sum_{k=x+1}^n binom{n}{k-1}p^k(1-p)^{n-k} }{left[sum_{k=x+1}^n binom{n}{k-1}p^k(1-p)^{n-k}right] +p^{n+1}(1-p)^{-1} }\
        &leq frac{n-x}{x+1}
        end{align}
        as desired.






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        $endgroup$



        We know that
        $$
        frac{P(X=x+1)}{P(X=x)}=frac{n-x}{x+1}frac{p}{1-p}
        $$
        Therefore, we will show that
        $$
        frac{P(Xgeq x+1)}{P(Xgeq x)}leq frac{n-x}{x+1}frac{p}{1-p}
        $$
        Multiplying the left hand side by $(1-p)/p$ and expanding out the numerator and denominator on the left hand side we have
        begin{align}
        frac{1-p}{p}frac{P(Xgeq x+1)}{P(Xgeq x)}&=frac{1-p}{p}frac{sum_{kgeq x+1}dbinom{n}{k}p^k(1-p)^{n-k}}{sum_{kgeq x}dbinom{n}{k}p^k(1-p)^{n-k}}\
        &=frac{binom{n}{x+1}p^{x+1}(1-p)^{n-x-1}+binom{n}{x+2}p^{x+2}(1-p)^{n-x-2}+binom{n}{x+3}p^{x+3}(1-p)^{n-x-3}+cdots}{binom{n}{x}p^{x+1}(1-p)^{n-x-1}+binom{n}{x+1}p^{x+2}(1-p)^{n-x-2}+binom{n}{x+2}p^{x+3}(1-p)^{n-x-3}+cdots}tag{1}
        end{align}



        Note that the numerator has $n-x$ terms while the denominator has $n-x+1$ terms. Now we examine ratios of successive binomial coefficients. Note that
        $$
        frac{binom{n}{x+1}}{binom{n}{x}}=frac{n-x}{x+1}
        $$
        and for $kgeq 1$
        $$
        frac{binom{n}{x+1+k}}{binom{n}{x+k}}=frac{n-x-k}{x+1+k}leq frac{n-x}{x+1}
        $$
        Combining the last two results, we may write for $kgeq 0$
        $$
        binom{n}{x+1+k}leq frac{n-x}{x+1}binom{n}{x+k}
        $$
        Substituting this inequality for each binomial coefficient in the numerator of (1), we have
        begin{align}
        frac{1-p}{p}frac{P(Xgeq x+1)}{P(Xgeq x)}&=frac{binom{n}{x+1}p^{x+1}(1-p)^{n-x-1}+binom{n}{x+2}p^{x+2}(1-p)^{n-x-2}+binom{n}{x+3}p^{x+3}(1-p)^{n-x-3}+cdots}{binom{n}{x}p^{x+1}(1-p)^{n-x-1}+binom{n}{x+1}p^{x+2}(1-p)^{n-x-2}+binom{n}{x+2}p^{x+3}(1-p)^{n-x-3}+cdots}\
        &leq frac{n-x}{x+1}frac{sum_{k=x+1}^n binom{n}{k-1}p^k(1-p)^{n-k} }{left[sum_{k=x+1}^n binom{n}{k-1}p^k(1-p)^{n-k}right] +p^{n+1}(1-p)^{-1} }\
        &leq frac{n-x}{x+1}
        end{align}
        as desired.







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        answered Jan 25 at 13:19









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