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I'm stuck with this convolution integral ($z \geq 0$)... \begin{equation} f_{Z}(z)=\int^{\infty}_{-\infty}f_{1}(x)f_{2}(z-x)dx = \mbox{ } ??? \end{equation} which represents the pdf of the sum $Z = X_1 + X_2$ of two random variables $X_1$ and $X_2$. Each $X_1$ and $X_2$ belongs to the same family (each of them is proportional to a non-central $\chi^2$ variable): \begin{equation} \begin{aligned} f_{1}(x) &= \alpha_1 e^{-\beta_1 x} (\gamma_1 x)^{\frac{1}{2} \nu} I_{\nu}(\delta_1 \sqrt{x}) \times \mathcal{I}_{(0,\infty)}(x)\\ f_{2}(x) &= \alpha_2 e^{-\beta_2 x} (\gamma_2 x)^{\frac{1}{2} \mu} I_{\mu}(\delta_2 \sqrt{x}) \times \mathcal{I}_{(0,\infty)}(x) \end{aligned} \end{equation} where $\mathcal{I}_{(0,\infty)}(\cdot)$ is the indicator function of the positive real axis and $I_{\nu}(\cdot)$ and $I_{\mu}(\cdot)$ are modified Bessel functions of the first kind of orders $\nu \geq 0$ and $\mu \geq 0$, respectively. Moreover, coefficients are all nonnegative: $\alpha_i \geq 0, \beta_i \geq 0, \gamma_i \geq 0, \delta_i \geq 0$.

My questions:

1) can anybody provide a closed-form expression for the integral? Off course I know I can easily use Fourier, but I'm really wondering whether such pdf can be explicitly write down.

2) Moreover, is it possible to write $f_{Z}(z)$ as $f_1$ and $f_2$, i.e. do (?) $\left\{\alpha_z, \beta_z, \gamma_z, \delta_z, q \right\}$ exist such that \begin{equation} f_{Z}(z) = \alpha_z e^{-\beta_z z} (\gamma_z z)^{\frac{1}{2} q} I_{q}(\delta_z \sqrt{z}) \end{equation}

3) I'm separately interested in the case: $\nu = \mu = 0$.

P.S. This question arises in this way: in a famous paper by Cox-Ingersoll-Ross (1985), there is a positive stochastic process $R(t)$ satisfying a square root SDE $$ dR = k(\theta - R)dt + \sigma \sqrt{R} dW $$ where $\alpha \geq 0$, $\theta \geq 0$, $\sigma \geq 0$ and $W_t$ is a standard Wiener process. Well, it can be shown that the conditional transition density (for time $s \geq t$) takes the same form above $$ f(R(s),s|R(t),t) = ce^{-u-v}\left(\frac{v}{u}\right)^{q/2}I_{q}(2\sqrt{uv}) $$ where \begin{equation} \begin{aligned} c &=\frac{2k}{\sigma^{2}\left(1-e^{-k(s-t)}\right)}\\ u &=cR(t)e^{-k(s-t)}\\ v &=cR(s)\\ q &=\frac{2k\theta}{\sigma^2}-1 \end{aligned} \end{equation} and $I_{q}(\cdot)$ is a modified Bessel function of the first kind of order $q$. Moreover it can be shown that: $2 c R(s) \sim \chi^2(2q + 2, 2u)$ i.e. the rescaled variable $2cR(s)$ distributes as a non central $\chi^2$ distribution with $2q+2$ degrees of freedom and parameter of non centrality $2u$. Note that, since the non central $\chi^2$ distribution is not closed w.r.t. the rescaling of the variable, $R(s)$ itself does not distribute according to a non central $\chi^2$. Therefore, my question is about the transitional distribution of the sum two independent processes $R_1$ and $R_2$ following \begin{equation} \begin{aligned} R &= R_1 + R_2 \\ dR_1 &= k_1(\theta_2 - R_1)dt + \sigma_1 \sqrt{R} dW_1 \\ dR_2 &= k_2(\theta_2 - R_2)dt + \sigma_2 \sqrt{R} dW_1 \\ \end{aligned} \end{equation} with $Corr[W_1, W_2] = 0$ and each having transitional density $f_{R_i}$. and therefore \begin{equation} f_{R}(r)=\int^{\infty}_{0}f_{R_1}(x)f_{R_2}(r-x)dx \end{equation} Off course $f_{R}(r)$ can be easily computed by numerical Fourier inversion, but as I said before, I would be courious about a closed form expression for the pdf.

Thanks for your attention.

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  • $\begingroup$ How to define $(z-x)^{1/2\mu}$ in this integral? $\endgroup$ – Sergei Jan 10 '15 at 21:14
  • $\begingroup$ Hi @Sergei, I don't understand your question, sorry.. $\endgroup$ – Gabriele Pompa Jan 10 '15 at 21:17
  • $\begingroup$ $z-x$ can be negative, its powers are complex numbers. For the integral it seems the only reasonable way is to use series for the second modified Bessel function and try after integration to sum up somehow. $\endgroup$ – Sergei Jan 10 '15 at 21:24
  • $\begingroup$ I'm sorry, I've been too sloppy in the notation. All the $x$ and $z$ are nonnegative. Therefore their pdf are 0 for $x<0$. However, I've found these pdfs have a name: "Bessel distributions of the first kind" (check for example Feller An Introduction to probability theory and its applications Vol.2). A particular case of these is the non central $\chi^2$. $\endgroup$ – Gabriele Pompa Jan 10 '15 at 21:56
  • $\begingroup$ So the integral in your question is $\int_0^z$? $\endgroup$ – Sergei Jan 11 '15 at 7:55

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