## gamma function – algebraic manipulation on equation

An equation involving the poisson point process is formulated as:

$$prod_{j=1}^{K}exp(-2pilambda_j(sp_j)^{2/alpha}int_0^{infty}rint_0^{infty}e^{-t(1+r^{alpha})}dtdr)$$.

Some algebraic manipulations are carried out and the equation is rewritten as:

$$exp(-s^{2/alpha}C(alpha)sum_{i=1}^{K}lambda_iP_i^{2/alpha})$$, where $$C(alpha)=frac{2pi csc(frac{2}{alpha})}{alpha}$$.

I only know the manipulation is related to Gamma function. I want to know the datails about the manipulation.

Thanks a lot! : )

## pr.probability – Inaccurate results for the analytical expression of \$mathbb{E}left[ a mathcal{Q} left( sqrt{b } gamma right) right]\$

I’m trying to plot a graph for the following expectation

$$mathbb{E}left( a mathcal{Q} left( sqrt{b } gamma right) right)=a 2^{-frac{kappa }{2}-1} b^{-frac{kappa }{2}} theta ^{-kappa } left(frac{, _2F_2left(frac{kappa }{2}+frac{1}{2},frac{kappa }{2};frac{1}{2},frac{kappa }{2}+1;frac{1}{2 b theta ^2}right)}{Gamma left(frac{kappa }{2}+1right)}-frac{kappa , _2F_2left(frac{kappa }{2}+frac{1}{2},frac{kappa }{2}+1;frac{3}{2},frac{kappa }{2}+frac{3}{2};frac{1}{2 b theta ^2}right)}{sqrt{2} sqrt{b} theta Gamma left(frac{kappa +3}{2}right)}right)$$
where $$a$$ and $$b$$ are constant values, $$mathcal{Q}$$ is the Gaussian Q-function, which is defined as $$mathcal{Q}(x) = frac{1}{sqrt{2 pi}}int_{x}^{infty} e^{-u^2/2}du$$ and $$gamma$$ is a random variable with Gamma distribition, i.e., $$f_{gamma}(y) sim frac{1}{Gamma(kappa)theta^{kappa}} y^{kappa-1} e^{-y/theta}$$ with $$kappa > 0$$ and $$theta > 0$$.

This equation was also found with Mathematica, so it seems to be correct.

Follows some examples, where I have checked the analytical results against the simulated ones.

When $$kappa = 12.85$$, $$theta = 0.533397$$, $$a=3$$ and $$b = 1/5$$ it returns the correct value $$0.0218116$$.

When $$kappa = 12.85$$, $$theta = 0.475391$$, $$a=3$$ and $$b = 1/5$$ it returns the correct value $$0.0408816$$.

When $$kappa = 12.85$$, $$theta = 0.423692$$, $$a=3$$ and $$b = 1/5$$ it returns the value $$-1.49831$$, which is negative. However, the correct result should be a value around $$0.0585$$.

When $$kappa = 12.85$$, $$theta = 0.336551$$, $$a=3$$ and $$b = 1/5$$ it returns the value $$630902$$. However, the correct result should be a value around $$0.1277$$.

Therefore, the issue happens as $$theta$$ decreases. For values of $$theta > 0.423692$$ the analytical matches the simulated results. The issue only happens when $$theta <= 0.423692$$.

I’d like to know if that is an accuracy issue or if I’m missing something here and if there is a way to correctly plot a graph that matches the simulation.

## calculus and analysis – Integral of \$r frac{2^{r-1} log (2) e^{-frac{sqrt{2^r-1}}{b}} left(2^r-1right)^{frac{d}{2}-1}}{b^d Gamma (d)}\$ with Mathematica

I’m trying to find the integral given below with Mathematica

$$int_0^{infty } r frac{2^{r-1} log (2) e^{-frac{sqrt{2^r-1}}{b}} left(2^r-1right)^{frac{d}{2}-1}}{b^d Gamma (d)} , dr$$

However, it takes too long for it to return something and when it returns it outputs the same integral.

$$int_{0}^{infty } frac{2^{r-1} r log (2) b^{-d} e^{-frac{sqrt{2^r-1}}{b}} left(2^r-1right)^{frac{d}{2}-1}}{Gamma (d)} , dr$$

I’d like to figure out the solution for this integral.

## Mathematica erroneously says this gamma integral does not converge

Mathematica does not recognize this case:

``````Integrate((T^a/(Gamma(a))) *  (E^(-Tx)) (x^(a - 1)), {x, 0, Infinity})
Integrate::idiv: Integral of (E^-Tx T^a x^(-1+a))/Gamma(a) does not converge on {0,(Infinity)}.
``````

This should converge because

## coding theory – Variance Gamma parameter estimation in R studio

I am using the vgFit function in R Studio to estimate parameters for stock prices. I converted the stock prices to a vector and I get this error. My code was:

``````param<-c(0,1,0,1)
fit<-vgFit(c(x), param=param)
``````

The error is:

``````Error in optim(paramStart, llsklp, NULL, method = startMethodSL, hessian = FALSE,  :
function cannot be evaluated at initial parameters
``````

## analytic number theory – Is every modular function on \$Gamma\$ univalent?

Suppose $$f$$ is a modular function on $$Gamma$$ then it has a fundamental region $$R_L$$. Since $$f$$ is modular, it can be expressed as a rational function of $$J(tau)=frac{g_2^3(tau)}{Delta(tau)}$$. To prove this we actually construct a rational function
$$g(tau)=frac{prod(J(a_i)-J(tau))}{prod(J(b_i)-J(tau))}$$
where the $$a_i,b_i$$ are zeros and poles of $$f$$. It seems to me that this construction implies the possibility of having multiple poles in the fundamental region. I know for sure that $$J$$ has only one pole since it attains every value exactly once.

Since a modular function is univalent if and only if its fundamental region has genus $$1$$ and if we identify the edges of $$R_Gamma$$ it would become homeomorphic to $$mathbb{P}^1$$, any modular function that has the fundamental region $$R_Gamma$$ should be univalent, that is, any modular function is univalent in the closure of its fundamental region.

This really confused me and I am not sure if this is correct.

## Fitting data to left skewed gamma distribution

How can I fit the following two set of data to a left skewed gamma function, which I what I think should fit the data best?:

data 1 is here: https://pastebin.com/X2HTjTP7
data 2 is here: https://pastebin.com/8Rh4BHDT

Is there any other suggestion of what would be the best distribution or equation to fit the data?

A picture of how data 1 looks is here:

A picture of how data 2 looks is here:

## simplifying expressions – Is there any way to prioritize Gamma function?

In Mathematica, if you type

``````Gamma[1, 0, -a]
``````

It automatically simplifies to `1 - E^a`. I wonder though if there is a way to stop Mathematica from simplify Gamma function automatically.

I understand that I can use `Inactive` or `HoldFrom`. But I actually want having to remember using Inactive every time.

## How to calculate gamma function values ​​for quaternions with Mathematica?

How can we calculate the values ​​for that? gamma Function with Quaternions on Mathematica?

For example:

``````Gamma(5) = 24

N(Gamma(I)) = -0.15495 - 0.498016 I

N(Gamma(1 + 2 I)) = 0.151904 + 0.0198049 I

N(Gamma(Quaternion(5, 0, 0, 0))) = Gamma(Quaternion(5., 0., 0., 0.)) (* which is wrong *)
``````

How can we calculate specifically?

``````Gamma(Quaternion(1, 2, 3, 4))?
``````

Tries:

``````N(Gamma(FromQuaternion(Quaternion(1, 2, 0, 0)))) (* works *)
N(Gamma(FromQuaternion(Quaternion(1, 2, 3, 4)))) (* doesn't work *)

QuaternionPower(x_, y_) := E^(y ** Log(x))
QuaternionGamma(z_) := Integrate(QuaternionPower(E, -x + Log(x)*(z - 1)), {x, 0, Infinity})
N(QuaternionGamma(FromQuaternion(Quaternion(1, 2, 0, 0)))) = 0.151904 + 0.0198049 I (* works *)
N(QuaternionGamma(FromQuaternion(Quaternion(1, 2, 3, 0)))) = ConditionalExpression(Gamma((1.  + 2. I) + 3. J), Re(J) > -0.333333) (* doesn't work *)
``````

## Logic – If a set of sentences \$ Gamma \$ has an infinite model, does it also have a denumerable model?

I think that's wrong, since we can only conclude from the Downward-Lowenheim-Skolem theorem $$Gamma$$ has an enumeration model, it is not shown whether this enumeration model is infinite. Certainly ours $$Gamma$$ already has an infinite model, but nothing says that its enumerable model must be identical to its infinite model.

I can't see how this claim is true, but I think it should be true?