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Inequality with known minimum values of parameters

Hi!

I want to show that when 0<p<1 then the inequality holds for all values of d and b that are >0. And for p>1 then the inequality does not hold for any values of d and b >0.

I know the answer as I have "solved" it numerically in excel, however I do not know how to do it with algebra. I have this inequality:

(1+(b*d)^(1/p))^(p-1) > (b*d+b)/((b*d)^(1/p)+b)

Please see attachment for picture of the inequality.

b, d and p are all greater than 0. It is clear that when p=1 than the inequality holds for all values of b and d.

It is for my thesis in economics. I am trying to show how the difference in consumption paths between two types of consumers depend on the parametric values.

Any help would be greatly appreciated!

Jacob

Re: Inequality with known minimum values of parameters

Hey Katmarn.

One recommendation I have is to differentiate the function with respect to the parameters (i.e. f(b,d,p) > 0) and use that to show that the derivatives of the function f have the right properties to follow (i.e. with respect to turning points and derivatives).

If you can for example show that you have monotonicity or a similar property over the range of the variables (like p) then you can show that the function is not negative.

For multi-variable scenarios you will need to look at the Hessian in the worst case and the single partial derivatives in the easiest case.

Since you have restrictions on your range, you can evaluate the appropriate derivatives at the extremities to see what the behavior is across the range.

Re: Inequality with known minimum values of parameters

Are you sure b and d can be be less than or equal to 1?

Take the case when bd=1, p=0.5

$\displaystyle (1+(bd)^{1/p})^{p-1}=(1+1^2)^{-0.5}=2^{-0.5}$

$\displaystyle \frac{bd+b}{bd^{1/p}+b}=\frac{1+b}{1+b}=1$

The inequality is not true in this case.

Re: Inequality with known minimum values of parameters

I have made a mistake - i'm very sorry!

The inequality sign should be turned around! I'm afraid I cannot edit in my original post.

Thank you

Re: Inequality with known minimum values of parameters

For any b,d>0 bd>0

so $\displaystyle 1+(bd)^{1/p}>1$

for 0<p<1 p-1 is negative so

$\displaystyle (1+(bd)^{1/p})^{p-1}<1$

That's the left hand side taken care of

Consider 3 cases for the value of bd

Case 1 bd=1

$\displaystyle \frac{bd+b}{bd^{1/p}+b}=\frac{1+b}{1^{1/p}+b}=1$

In case 1 the right hand side is equal to 1 and the left hand side is less than 1 so the inequality holds

Case 2 bd<1

$\displaystyle \frac{1}{p}>1$ so it is clear to see that

$\displaystyle (bd)^{1/p}<bd$ Therefore, $\displaystyle (bd)^{1/p}+b<bd+b$

And $\displaystyle \frac{bd+b}{bd^{1/p}+b}>1$

In case 2 the right hand side is greater than 1 and the left hand side is less than 1 so the inequality holds

Case 3 bd>1

$\displaystyle \frac{1}{p}>1$ so it is clear to see that

$\displaystyle (bd)^{1/p}>bd$ Therefore, $\displaystyle (bd)^{1/p}+b>bd+b$

And $\displaystyle \frac{bd+b}{bd^{1/p}+b}>1$

In case 3 it is not clear if the inequality is true so if you continue trying to prove the inequality you need only examine case 3.

Re: Inequality with known minimum values of parameters

I will look into it. Thank you very much for your help!

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Re: Inequality with known minimum values of parameters

Shakarri, I have another question that looks a lot like the first one. I just can't seem to "see the light".

Again, how to show that for 0<p<1, 0<b<1 and d>0 that:

(1+(b*d)^{1/p})^{p-1}*(1+d*(b*d)^{(1-p)/p})<(1+d^{1/p})^{p}

And that for p>1, 0<b<1 and d>0 that:

(1+(b*d)^{1/p})^{p-1}*(1+d*(b*d)^{(1-p)/p})>(1+d^{1/p})^{p}

I hope you (or someone else) have the time to help.

Best

Jacob

Re: Inequality with known minimum values of parameters

I was able to do a but more of the first question.

Continuing case 3 where bd>1

Let bd=k

$\displaystyle ((bd)^{\frac{1}{p}}+1)^{p-1}<\frac{bd+b}{(bd)^{\frac{1}{p}}+b}$

$\displaystyle (k^{\frac{1}{p}}+1)^{p-1}<\frac{k+b}{k^{\frac{1}{p}}+b}$

We seek to find the value of b which minimises the equation. For any values of k and p the minimum of $\displaystyle \frac{k+b}{k^{\frac{1}{p}}+b}$

Occurs at $\displaystyle \frac{d}{db}(\frac{k+b}{k^{\frac{1}{p}}+b})=0$

Doing the differentiation $\displaystyle \frac{k^{\frac{1}{p}}+b-(k+b)}{(k^{\frac{1}{p}}+b)^2}=0$

$\displaystyle \frac{k^{\frac{1}{p}}-k}{(k^{\frac{1}{p}}+b)^2}=0$

$\displaystyle k^{\frac{1}{p}}-k=0$

The equation has no absolute minimum however you can find a local minimum for b>0

Because k>1 and 0<p<1, $\displaystyle k^{\frac{1}{p}}<k$

The slope $\displaystyle \frac{k^{\frac{1}{p}}-k)}{(k^{\frac{1}{p}}+b)^2}$ is always positive therefore the minimum value of $\displaystyle \frac{k+b}{k^{\frac{1}{p}}+b}$ occurs at the minimum value of b. b has no minimum, instead the minimum will occur in the limit as b tends to zero.

$\displaystyle lim_{b\rightarrow 0}\frac{k+b}{k^{\frac{1}{p}}+b}=\frac{k}{k^{\frac{ 1}{p}}}$

Note that even though k=bd, in the limit as b tends to zero k does not have to be zero as there is always a value of d which can make k still be greater than 1.

Since $\displaystyle \frac{k}{k^{\frac{1}{p}}}$ is the minimum of $\displaystyle \frac{k+b}{k^{\frac{1}{p}}+b}$ we know that $\displaystyle \frac{k}{k^{\frac{1}{p}}}<\frac{k+b}{k^{\frac{1}{p }}+b}$

If we can show that $\displaystyle ((k)^{\frac{1}{p}}+1)^{p-1} < \frac{k}{k^{\frac{1}{p}}}$ then we know that $\displaystyle ((k)^{\frac{1}{p}}+1)^{p-1}<\frac{k+b}{(k)^{\frac{1}{p}}+b}$

I do not know how to prove this final step, however you could do a 3D plot of $\displaystyle ((k)^{\frac{1}{p}}+1)^{p-1} - \frac{k}{k^{\frac{1}{p}}}$ and see if at any point it becomes less than zero. This is not a formal proof but it might suffice for an economics thesis. A plot is also limited to how large you can make it but I assume you know some reasonable maximum value of d and in reality you don't find values of d all the way up to infinity.

Re: Inequality with known minimum values of parameters

Nevermind you wont have to do a 3D plot :D

Continuing from $\displaystyle (k^{\frac{1}{p}}+1)^{p-1}<\frac{k}{k^{\frac{1}{p}}}$

$\displaystyle (k^{\frac{1}{p}}+1)^{p-1}<k^1\times k^{-\frac{1}{p}}$

$\displaystyle (k^{\frac{1}{p}}+1)^{p-1}<k^{\frac{p-1}{p}}$

Since p-1 is negative when I put both sides to the power of 1/(p-1) the inequality changes direction

$\displaystyle (k^{\frac{1}{p}}+1)>k^{\frac{1}{p}}$

$\displaystyle 1>0$

Re: Inequality with known minimum values of parameters

The second part of your first post, to show that the inequality is false for p>1 works much the same.

For any b,d>0 bd>0

so $\displaystyle 1+(bd)^{1/p}>1$

for p>1 p-1 is positive so

$\displaystyle (1+(bd)^{1/p})^{p-1}>1$

That's the left hand side taken care of

Consider 3 cases for the value of bd

Case 1 bd=1

$\displaystyle \frac{bd+b}{(bd)^{1/p}+b}=\frac{1+b}{1^{1/p}+b}=1$

In case 1 the right hand side is equal to 1 and the left hand side is greater than 1 so the inequality is false

Case 2 bd<1

$\displaystyle \frac{1}{p}<1$ so it is clear to see that

$\displaystyle (bd)^{1/p}>bd$ Therefore, $\displaystyle (bd)^{1/p}+b>bd+b$

And $\displaystyle \frac{bd+b}{(bd)^{1/p}+b}<1$

In case 2 the right hand side is less than 1 and the left hand side is greater than 1 so the inequality is false

Case 3 bd>1

$\displaystyle \frac{1}{p}>1$ so it is clear to see that

$\displaystyle (bd)^{1/p}<bd$ Therefore, $\displaystyle (bd)^{1/p}+b<bd+b$

And $\displaystyle \frac{bd+b}{bd^{1/p}+b}>1$

In case 3 it is not clear if the inequality is false or not.

Now it gets a bit different.

To show that in case 3 when bd>1 and p>1 the inequality is false I will just show that it is false for $\displaystyle p\geq 1$. Again I'm using k=bd to make it easier to write out.

Consider the inequality

$\displaystyle (1+k^{1/p})^{p-1}>\frac{k+b}{k^{1/p}+b}$

Let $\displaystyle f(p)=(1+k^{1/p})^{p-1}$ and $\displaystyle g(p)=\frac{k+b}{k^{1/p}+b}$

When p=1 f(p)=1 and g(p)=1, thus for p=1 the inequality is false.

If the inequality $\displaystyle f(p)<g(p)$ is to remain false for all $\displaystyle p\geq 1$ f(p) must grow faster than g(p). If f(p) begins (at p=1) greater than or equal to g(p) and grows at a faster rate than f(p) then g(p) will always be greater than or equal to f(p).

Thus we must show that $\displaystyle g'(p)\geq f'(p)$ This is the monotonicity which chiro was talking about

The rate f(p) grows at is its derivative, I wont show the differentiation here but $\displaystyle f'(p)=(p-1)(k^{\frac{1}{p}}+1)^{p-2}k^{\frac{1}{p}}ln(k)(-p^{-2})$

And $\displaystyle g'(p)=\frac{-(k+b)k^{\frac{1}{p}}ln(k)(-p^{-2})}{(k^{\frac{1}{p}}+b)^2}$

So the inequality is false if $\displaystyle g'(p)=\frac{-(k+b)k^{\frac{1}{p}}ln(k)(-p^{-2})}{(k^{\frac{1}{p}}+b)^2} \geq (p-1)(k^{\frac{1}{p}}+1)^{p-2}k^{\frac{1}{p}}ln(k)(-p^{-2})$

Divide both sides by $\displaystyle k^{\frac{1}{p}}ln(k)p^{-2})$

$\displaystyle \frac{-(k+b)(-1)}{(k^{\frac{1}{p}}+b)^2} \geq (p-1)(k^{\frac{1}{p}}+1)^{p-2}(-1)$

$\displaystyle \frac{(k+b)}{(k^{\frac{1}{p}}+b)^2} \geq (p-1)(k^{\frac{1}{p}}+1)^{p-2}(-1)$

The left hand side is positive and the right hand side is negative, therefore f(p)<g(p) for all $\displaystyle p \geq 0$

Could you tell me the title of your thesis? I'd like to look at it when it is finished.

Re: Inequality with known minimum values of parameters

Thank you so much for your big help!

The title is: Short run discounting: Intra-monthly evidence from Danish bank statements

All the best

Re: Inequality with known minimum values of parameters

Doing the first two was really enjoyable (Nod)

The inequalities in your second question are more difficult to prove. Have you already "proven" them in excel?

Also for the last part when you are trying to show that http://puu.sh/3r8j6.jpg holds when p>1 are you sure the inequality isn't meant to be $\displaystyle \geq$? If you are just trying to show that http://puu.sh/3r8qQ.jpg is false then it's possible for the two sides to be equal.

Re: Inequality with known minimum values of parameters

I noticed that the first equation in my post above is untrue for small values of bd. For example, b=0.2, d=0.5, p=0.6 I hope this doesn't interfere with your theory too much.

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Re: Inequality with known minimum values of parameters

Hi Shakarri!

I'm glad you enjoyed the first two :)

I have "proven" the second question in Excel - please see the attached excel file in sheet 2.

When p=1 the left and right hand side are equal (an important result for me). This is easy to see as p=1 reduces the inequalities to:

(1+bd)^0*(1+d*(bd)^0)>(1+d) = 1+d>1+d

Please remember the restrictions: 0<b<1 and 0<d<=1 ( made the restriction on d a little harder as is does not have economic meaning if d>1).

If you are interested we are showing whether a sophisticated hyperbolic agent consumes more or less in the first period than a naïve hyperbolic agent. d is the standard discount factor and b is the hyperbolic discount factor.

See Laibson's article "Golden eggs and hyperbolic discounting" for more on Quasi-hyperbolic discounting. Scholar link: Golden eggs and hyperbolic discounting - Google Scholar

Re: Inequality with known minimum values of parameters

That extra restraint on d will be useful. You got the inequality sign backwards again, that's why it was so hard to prove!

This proof contains a lot of cumbersome partial differentiation, it is long winded but it does the job.

Once again let bd=k

as $\displaystyle 0<d\leq 1$ and $\displaystyle 0<b<1$ we know that $\displaystyle 0<k<1$, and $\displaystyle 0<p<1$

The inequality to prove is

$\displaystyle (1+k^\frac{1}{p})^{p-1}(1+dk^\frac{1-p}{p})<(1+d^{\frac{1}{p})^p$

Let $\displaystyle f(d,k,p)=(1+k^\frac{1}{p})^{p-1}(1+dk^\frac{1-p}{p})$ and $\displaystyle g(d,p)=(1+d^{\frac{1}{p}})^p$

Now Let $\displaystyle h(d,k,p)=\frac{f(d,k,p)}{g(d,p)}$

Prove that $\displaystyle h(d,k,p)<1$

I will be using the method which Chiro suggested.

First step, given any values of p and d find the value of b which maximises h(d,k,p)

The differentiation is easier when using k instead of b so I'll do the differentiation in terms of k then convert back to the variable b.

Suppose k^{*} is the value of k which maximises h(d,k,p). If we find k^{*} then we know that h(d,k,p)<h(d,k^{*},p) and if we then prove that h(d,k^{*},p)<1 we know that h(d,k,p)<1 for all k

The rate of change of h with respect to k is $\displaystyle \frac{\partial h}{\partial k}$

From $\displaystyle h(d,k,p)=\frac{(1+k^\frac{1}{p})^{p-1}(1+dk^\frac{1-p}{p})}{(1+d^{\frac{1}{p}})^p}$ Let $\displaystyle u= (1+k^\frac{1}{p})^{p-1}, v=(1+dk^\frac{1-p}{p})$

$\displaystyle h(d,k,p)=\frac{uv}{(1+d^{\frac{1}{p}})^p}$

$\displaystyle \frac{\partial h}{\partial k}=\frac{1}{(1+d^{\frac{1}{p}})^p}[(1+k^\frac{1}{p})^{p-1}\frac{\partial v}{\partial k}+(1+dk^\frac{1-p}{p}) \frac{\partial u}{\partial k}]$

$\displaystyle u= (1+k^\frac{1}{p})^{p-1}, a_1=1+k^\frac{1}{p}, u=a_1^{p-1}$

$\displaystyle \frac{\partial u}{\partial k}=\frac{\partial u}{\partial a_1}\frac{\partial a_1}{\partial k}=(p-1)a_1^{p-2}(\frac{1}{p} k^{\frac{1}{p}-1})=\frac{p-1}{p}(1+k^{\frac{1}{p}})^{p-2}k^{\frac{1-p}{p}}$

$\displaystyle v=(1+dk^\frac{1-p}{p})$

$\displaystyle \frac{\partial v}{\partial k}=d\frac{1-p}{p}k^{\frac{1-p}{p}-1}=d\frac{1-p}{p}k^{\frac{1}{p}-2}$

$\displaystyle \frac{\partial h}{\partial k}=\frac{1}{(1+d^{\frac{1}{p}})^p}[(1+k^\frac{1}{p})^{p-1}d\frac{1-p}{p}k^{\frac{1}{p}-2}+(1+dk^\frac{p-1}{p}) \frac{p-1}{p}(1+k^{\frac{1}{p}})^{p-2}k^{\frac{1-p}{p}}]$

Factorising out a lot

$\displaystyle \frac{\partial h}{\partial k}= \frac{1}{(1+d^{\frac{1}{p}})^p} \frac{1-p}{p} (1+k^\frac{1}{p})^{p-1}k^{\frac{1-p}{p}}[\frac{d}{k}+\frac{-1-dk^{\frac{1-p}{p}}}{1+k^\frac{1}{p}}]$

$\displaystyle \frac{\partial h}{\partial k}= \frac{1}{(1+d^{\frac{1}{p}})^p} \frac{1-p}{p} (1+k^\frac{1}{p})^{p-1}k^{\frac{1-p}{p}}[\frac{d(1+k^\frac{1}{p})-k(1+dk^{\frac{1-p}{p}})}{k(1+k^\frac{1}{p})}]$

This simplifies nicely to

$\displaystyle \frac{\partial h}{\partial k}= \frac{1}{(1+d^{\frac{1}{p}})^p} \frac{1-p}{p} (1+k^\frac{1}{p})^{p-1}k^{\frac{1-p}{p}}[\frac{d-k}{k(1+k^\frac{1}{p})}]$

Converting back to the variable b

$\displaystyle k=bd$

$\displaystyle \frac{\partial k}{\partial b}=d$

$\displaystyle \frac{\partial h}{\partial b}=\frac{\partial h}{\partial k}\frac{\partial k}{\partial b}$

$\displaystyle \frac{\partial h}{\partial b}=\frac{\partial h}{\partial k}d$

$\displaystyle \frac{\partial h}{\partial b}= d\frac{1}{(1+d^{\frac{1}{p}})^p} \frac{1-p}{p} (1+k^\frac{1}{p})^{p-1}k^{\frac{1-p}{p}}[\frac{d-k}{k(1+k^\frac{1}{p})}]$

$\displaystyle \frac{\partial h}{\partial b}= d\frac{1}{(1+d^{\frac{1}{p}})^p} \frac{1-p}{p} (1+(bd)^\frac{1}{p})^{p-1}(bd)^{\frac{1-p}{p}}[\frac{d-(bd)}{(bd)(1+(bd)^\frac{1}{p})}]$

If you look at all the parts which multiply together to give the full expression you will see that each part is positive. The slope is positive so the maximum value of h(d,k,p) occurs at the max value of b. b has no maximum value but take the case when b=1

When b=1 $\displaystyle h(d,k,p)=\frac{(1+d^\frac{1}{p})^{p-1}(1+d^\frac{1}{p})}{(1+d^{\frac{1}{p}})^p}=1$

Because the slope is positive (and never zero) for any value of b less than 1, h(d,k,p) will be less than the value of h(d,k,p) when b=1. Therefore for b<1, h(d,k,p)<1 and this proves the original inequality.