Kurtosis of made up distribution Take a look at the image below. Blue line indicates standard normal pdf. The red zone is supposed to be equal to the sum of grey areas (sorry for awful drawing).
I wonder can we create a new distribution with higher peak by shifting grey zones to the top (red zone) of the normal pdf?


If such transformation can be made, than what do you think about the kurtosis of this new distribution? Leptokurtic? But it has the same tails as the normal distribution does! Undefined?
 A: Kurtosis is a rather misunderstood concept (I find L.T. De Carlo's paper "On the Meaning and Use of Kurtosis" (1997) a sensible and valuable discussion and presentation of the issues involved).  
So I will take the naive view, and I will construct a density, $g_X(x)$, with "thinner middle and higher value at mode", compared to the standard normal density, but identical "tails" with the latter. I do not claim that this density exhibits "excess kurtosis".  
This density will necessarily be step-wise. In order to have identical left and right "tails", its functional form for the intervals $(-\infty, -a)$ and $(a,\infty)$, where $a>0$, should be identical to the standard normal $\phi(x)$ density. 
In the middle interval, $(-a,a)$, it should have some other functional form, call it $h(x)$. This $h(x)$ should be symmetric around zero, and satisfy   
1) $h(0) > \phi(0) = 1/\sqrt{2\pi}$ so that the value of the density at the mode will be higher than the value of the standard normal, and   
2) $\phi(-a) = h(-a) = h(a) = \phi(a)$ so that $g_X(x)$ is continuous.  
More over, $g_X(x)$ should integrate to unity over the domain, in order to be a proper density.
So this density will be
$$g_X(x) =  \begin{matrix}
\phi(x) &-\infty<x\le -a\\
h(x) &-a\le x \le a\\
\phi(x) & a\le x<\infty 
\end{matrix}$$
subject to the previously mentioned restrictions on $h(x)$ and also, subject to  
$$\int_{-\infty}^{-a}\phi(t)dt + \int_{-a}^ah(t)dt + \int_{a}^{\infty}\phi(t)dt =1$$
which is equivalent to require that the probability mass under $h(x)$ in the interval $(-a,a)$ must be equal with the probability mass under $\phi(x)$ in the same interval:
$$\int_{-a}^{-a}\left(h(t)- \phi(t)\right)dt =0 \Rightarrow  \int_{0}^{a}\left(h(t)- \phi(t)\right)dt=0 $$
the last part due to the symmetry properties.  
To obtain something specific, we will "try" the density of the zero-mean Laplace distribution for $h(x)$ 
$$h(x)= \frac 1{2b} e^{-\frac {|x|}{b}},\; b>0$$
To satisfy the various requirements set previously we must have:
For higher value at mode,
$$h(0)= \frac 1{2b} > \phi(0) = \frac {1}{\sqrt{2\pi}} \Rightarrow 0<b < \sqrt{\pi/2} \qquad [1]$$
For continuity,
$$h(a) = \phi(a) \Rightarrow \frac 1{2b} e^{-\frac {a}{b}} = \frac {1}{\sqrt {2\pi}}e^{-\frac 12a^2}$$
$$\Rightarrow -\ln(2b) - \frac {a}{b} = -\ln(\sqrt {2\pi})  -\frac 12a^2 \Rightarrow \frac 12a^2 - \frac {a}{b} +\ln\frac{\sqrt {\pi/2}}{b}$$
This is a quadratic in $a$. Its discriminant is 
$$\Delta_a = \frac 1{b^2} - 4\cdot \frac 12 \cdot\ln\frac{\sqrt {\pi/2}}{b} > 0$$
(it can be easily verified that it is always positive). More over, we keep only the positive root since $a>0$ so
$$a^* = \frac 1b + \sqrt{\Delta_a}\qquad [2]$$
Finally the requirement for the density to integrate to unity translates into
$$\int_{0}^{a^*}\frac 1{2b} e^{-\frac {|x|}{b}} dt = \int_{0}^{a^*}\phi(t)dt $$
which by straightforward integration leads to
$$1-e^{-\frac {a^*}{b}} = 2\left(\Phi(a^*) - \frac 12\right) = \operatorname{erf}(a^*/\sqrt2)\qquad [3]$$
which can be solved numerically for $b^*$, and so completely determine the density we are after.  
Of course other functional forms symmetric around zero could be tried, the laplacian pdf was just for expositional purposes.  
A: The kurtosis of this distribution will probably be higher than that of a normal distribution.  I say probably because I am basing this on a rough drawing, and although it might be possible to prove that moving mass in this way always increases kurtosis, I am not positive about that.
Although it is true that it has the same tails as a normal distribution, this distribution will have a lower variance than the normal distribution from which it is derived.  Which means that its tails will match the tails of some normal distribution, but not of a normal distribution with the same variance as it.  So, the normalized tails will in fact be thicker than the tails of a normal distribution.  And, although thicker tails does not automatically mean more kurtosis, in this case the normalized fourth moment will probably also be larger.
A: It looks like the OP is trying to establish a connection between "peakedness" and kurtosis by keeping the tails fixed and making the distribution more "peaked." There is an effect on kurtosis here, but it is so slight that it is hardly worth a mention. Here is a theorem to support that assertion. 
Theorem 1: Consider any probability distribution with finite fourth moment. Construct a new probability distribution by replacing the mass in the $[\mu - \sigma, \mu + \sigma] $ range, keeping the mass outside of $[\mu - \sigma, \mu + \sigma] $  fixed, and keeping the mean and standard deviation at $\mu, \sigma$.  Then the difference between the minimum and maximum Pearson moment kurtosis values over all such replacements is $\le 0.25$. 
Comment: The proof is constructive; you can actually identify the min and max kurtosis replacements in this setting. Further, 0.25 is an upper bound on the kurtosis range, depending on the distribution. For example, with a normal distribution, the range bound is 0.141, rather than 0.25.
On the other hand, there is a huge effect of tails on kurtosis, as is given by the following theorem:
Theorem 2: Consider any probability distribution with finite fourth moment. Construct a new probability distribution by replacing the mass outside the $[\mu - \sigma, \mu + \sigma] $ range, keeping the mass in $[\mu - \sigma, \mu + \sigma] $ fixed, and keeping the mean and standard deviation at $\mu, \sigma$. Then the difference between the minimum and maximum Pearson moment kurtosis values over all such replacements is unbounded; i.e., the new distribution can be chosen so that kurtosis is aribitrarily large.
Comment: These two theorems show that the effect of tails on Pearson moment kurtosis is infinite, while the effect of "peakedness" is $\le 0.25$.
