Scaling a wavelet in continuous wavelet transform The mexican hat wavelet, obtained from the second derivative of a Gaussian, has a functional form of 

$$ \frac{2}{\sqrt{3 \sqrt{\pi} \sigma \ }} \left(1-\frac{t^2}{\sigma^2} \right) e^{-t^2/(2\sigma^2)} $$

where "$\sigma$ is the standard deviation of a Gaussian and $t$ is the independent variable.

If we graphically wish to graphically see its dilations, say at scale $1 \le a \le 5$, the typical form of a wavelet function $\psi$ has a scale parameter $a$ and a translation parameter $b$. The functional form of the Mexican hat given in MATLAB does not explicitly have them both.

The standard deviation $\sigma$ is the only variable here that controls with width, but this is not strictly equal to the scale $a$. How should we generate scaled versions of the Mexican hat?

Thanks.
Okay, I am not quite understanding your question.  Let's start with the definition of the Gaussian, aka the Bell Curve, in its general form.

$$  f(t) = \frac{1}{ \sigma \sqrt{2\pi}} e^{ -\frac{(t-\mu)^2}{2\sigma^2} } $$

$\mu$ is the mean, and represents where the peak occurs.

$\sigma$ is the standard deviation, and identifies where the inflection points are.

Taking the first derivative:

$$
\begin{aligned}
f'(t) &amp; = \frac{1}{ \sigma \sqrt{2\pi}}\left[ e^{ -\frac{(t-\mu)^2}{2\sigma^2} } \left( -\frac{(t-\mu)}{\sigma^2} \right) \right] \\
 &amp; = - \frac{1}{ \sigma^3 \sqrt{2\pi}}\left[ e^{ -\frac{(t-\mu)^2}{2\sigma^2} } \left( t-\mu \right) \right] \\
\end{aligned}
$$

Taking the second derivative.

$$
\begin{aligned}
f''(t) &amp;= - \frac{1}{ \sigma^3 \sqrt{2\pi}}\left[  e^{ -\frac{(t-\mu)^2}{2\sigma^2} } \left( -\frac{(t-\mu)}{\sigma^2} \right) \left( t-\mu \right) +
e^{ -\frac{(t-\mu)^2}{2\sigma^2} }   \left( 1 \right)  \right] \\
f''(t) &amp;=  \frac{1}{ \sigma^3 \sqrt{2\pi}}\left[    \frac{(t-\mu)^2}{\sigma^2} - 1 \right]  e^{ -\frac{(t-\mu)^2}{2\sigma^2} } \\
f''(t) &amp;=  \frac{1}{ \sigma^3 \sqrt{2\pi}}\left[  \left( \frac{t-\mu}{\sigma} \right)^2 - 1 \right]  e^{ - \frac{1}{2} \left( \frac{t-\mu}{\sigma} \right)^2 } \\
\end{aligned}
$$

In your example $\mu=0$ and $\sigma=1$.  In your graph, the zero is where the peak is, and the one is the +/-1 on your horizontal scale.  Notice that the second derivative is zero at these points (inflection in the original function.)

You can translate the graph by setting $\mu$.  You can stretch the graph horizontally by setting $\sigma$.

I am not familiar with wavelets, so I don't know what this $a$ and $b$ are that you are talking about.  You can definitely to another coordinate conversion if you want:

$$ t = a \tau + b $$

Gosh, but that is awfully similar to 

$$ \frac{1}{\sigma} t - \frac{\mu}{\sigma} $$

that you already got.



Based on the comments, to retrofit the function call to MATLAB (or wherever):

$$ \mu = b $$

$$ \sigma = 0.63628 a $$