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William
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I'm interested in finding the CDF and PDF of $U_i$ defined as follows, $$U_i=\frac g{d^{\alpha}}$$ where $g$ is a gamma distributed random variable with shape $k$ and scale $\theta$, and $d$ is a random variable with distribution $f_d(x)=\frac {2x}{R^2}, 0\le x \le R$.
To find the CDF of $U_i$ $$F_{U_i}(y)=P(U_i<y)=\int_0^R F_g(yx^\alpha)dx$$$$F_{U_i}(y)=P(U_i<y)=\int_0^R F_g(yx^\alpha)f_d(x)dx$$ where $F_g(y)$ is the CDF of gamma distributed random variable.
After performing the integration, which requires the use of the expansion of incomplete gamma function, I came to the following CDF: $$F_{U_i}(y)=\sum_{n=0}^\infty \frac {2(-1)^nR^{k\alpha+n\alpha}y^{k+n}}{\Gamma(k)n!\theta^{k+n}(k+n)(k\alpha+n\alpha+2)}$$ I want the CDF in this format because my ultimate goal is to find the CDF of $U=\sum_{i=1}^NU_i$ which requires me to find the Laplace transform of $F_{U_i}(y)$ and raise it to power $N$.
Now it's easy to find the PDF of $U_i$, i.e. $f_{U_i}(y)=\frac {d}{dy}F_{U_i}(y)$.
I'm looking for someone to clarify these points to me:

  • My derived $f_{U_i}(y)$ doesn't make sense to me, because it's based on a divergent series. How can it's integration from zero to infinity equals one?
  • In Matlab I took the $K^{th}$ elements of the series, and plotted it along with the histogram of 1000 $U_i$'s, the two plots are not related.
  • If you look at the definition of $U_i$, $U_i$ can take values from 0 to infinity, high values take less probability. This fact is not clear in the derived distribution $f_{U_i}(y)$. What is the domain of the derived PDF? What is the integration limit if I wannna find $F_\epsilon(\epsilon)=\int f(y)F_{U_i}(\epsilon(y+\sigma^2))dy$? where $f(y)$ is another density function.
  • How can I simulate the derived PDF in Matlab to check if it's really correlated with $U_i$

    • I'm interested in finding the CDF and PDF of $U_i$ defined as follows, $$U_i=\frac g{d^{\alpha}}$$ where $g$ is a gamma distributed random variable with shape $k$ and scale $\theta$, and $d$ is a random variable with distribution $f_d(x)=\frac {2x}{R^2}, 0\le x \le R$.
    To find the CDF of $U_i$ $$F_{U_i}(y)=P(U_i<y)=\int_0^R F_g(yx^\alpha)dx$$ where $F_g(y)$ is the CDF of gamma distributed random variable.
    After performing the integration, which requires the use of the expansion of incomplete gamma function, I came to the following CDF: $$F_{U_i}(y)=\sum_{n=0}^\infty \frac {2(-1)^nR^{k\alpha+n\alpha}y^{k+n}}{\Gamma(k)n!\theta^{k+n}(k+n)(k\alpha+n\alpha+2)}$$ I want the CDF in this format because my ultimate goal is to find the CDF of $U=\sum_{i=1}^NU_i$ which requires me to find the Laplace transform of $F_{U_i}(y)$ and raise it to power $N$.
    Now it's easy to find the PDF of $U_i$, i.e. $f_{U_i}(y)=\frac {d}{dy}F_{U_i}(y)$.
    I'm looking for someone to clarify these points to me:

  • My derived $f_{U_i}(y)$ doesn't make sense to me, because it's based on a divergent series. How can it's integration from zero to infinity equals one?
  • In Matlab I took the $K^{th}$ elements of the series, and plotted it along with the histogram of 1000 $U_i$'s, the two plots are not related.
  • If you look at the definition of $U_i$, $U_i$ can take values from 0 to infinity, high values take less probability. This fact is not clear in the derived distribution $f_{U_i}(y)$. What is the domain of the derived PDF? What is the integration limit if I wannna find $F_\epsilon(\epsilon)=\int f(y)F_{U_i}(\epsilon(y+\sigma^2))dy$? where $f(y)$ is another density function.
  • How can I simulate the derived PDF in Matlab to check if it's really correlated with $U_i$

    • I'm interested in finding the CDF and PDF of $U_i$ defined as follows, $$U_i=\frac g{d^{\alpha}}$$ where $g$ is a gamma distributed random variable with shape $k$ and scale $\theta$, and $d$ is a random variable with distribution $f_d(x)=\frac {2x}{R^2}, 0\le x \le R$.
    To find the CDF of $U_i$ $$F_{U_i}(y)=P(U_i<y)=\int_0^R F_g(yx^\alpha)f_d(x)dx$$ where $F_g(y)$ is the CDF of gamma distributed random variable.
    After performing the integration, which requires the use of the expansion of incomplete gamma function, I came to the following CDF: $$F_{U_i}(y)=\sum_{n=0}^\infty \frac {2(-1)^nR^{k\alpha+n\alpha}y^{k+n}}{\Gamma(k)n!\theta^{k+n}(k+n)(k\alpha+n\alpha+2)}$$ I want the CDF in this format because my ultimate goal is to find the CDF of $U=\sum_{i=1}^NU_i$ which requires me to find the Laplace transform of $F_{U_i}(y)$ and raise it to power $N$.
    Now it's easy to find the PDF of $U_i$, i.e. $f_{U_i}(y)=\frac {d}{dy}F_{U_i}(y)$.
    I'm looking for someone to clarify these points to me:

  • My derived $f_{U_i}(y)$ doesn't make sense to me, because it's based on a divergent series. How can it's integration from zero to infinity equals one?
  • In Matlab I took the $K^{th}$ elements of the series, and plotted it along with the histogram of 1000 $U_i$'s, the two plots are not related.
  • If you look at the definition of $U_i$, $U_i$ can take values from 0 to infinity, high values take less probability. This fact is not clear in the derived distribution $f_{U_i}(y)$. What is the domain of the derived PDF? What is the integration limit if I wannna find $F_\epsilon(\epsilon)=\int f(y)F_{U_i}(\epsilon(y+\sigma^2))dy$? where $f(y)$ is another density function.
  • How can I simulate the derived PDF in Matlab to check if it's really correlated with $U_i$
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    William
    William
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    William
    William
    • 127
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    I'm interested in finding the CDF and PDF of $U_i$ defined as follows, $$U_i=\frac g{d^{\alpha}}$$ where $g$ is a gamma distributed random variable with shape $k$ and scale $\theta$, and $d$ is a random variable with distribution $f_d(x)=\frac {2x}{R^2}, 0\le x \le R$.
    To find the CDF of $U_i$ $$F_{U_i}(y)=P(U_i<y)=\int_0^R F_g(yx^\alpha)dx$$ where $F_g(y)$ is the CDF of gamma distributed random variable.
    After performing the integration, which requires the use of the expansion of incomplete gamma function, I came to the following CDF: $$F_{U_i}(y)=\sum_{n=0}^\infty \frac {2(-1)^nR^{k\alpha+n\alpha}y^{k+n}}{\Gamma(k)n!\theta^{k+n}(k+n)(k\alpha+n\alpha+2)}$$ I want the CDF in this format because my ultimate goal is to find the CDF of $U=\sum_{i=1}^NU_i$ which requires me to find the Laplace transform of $F_{U_i}(y)$ and raise it to power $N$.
    Now it's easy to find the PDF of $U_i$, i.e. $f_{U_i}(y)=\frac {d}{dy}F_{U_i}(y)$.
    I'm looking for someone to clarify these points to me:

  • My derived $f_{U_i}(y)$ doesn't make sense to me, because it's based on a divergent series. How can it's integration from zero to infinity equals one?
  • In Matlab I took the $K^{th}$ elements of the series, and plotted it along with the histogram of 1000 $U_i$'s, the two plots are not related.
  • If you look at the definition of $U_i$, $U_i$ can take values from 0 to infinity, high values take less probability. This fact is not clear in the derived distribution $f_{U_i}(y)$. What is the domain of the derived PDF?   What is the integration limit if I wannna find $F_\epsilon(\epsilon)=\int f(y)F_{U_i}(\epsilon(y+\sigma^2))dy$? where $f(y)$ is another density function.
  • How can I simulate the derived PDF in Matlab to check if it's really correlated with $U_i$

    • I'm interested in finding the CDF and PDF of $U_i$ defined as follows, $$U_i=\frac g{d^{\alpha}}$$ where $g$ is a gamma distributed random variable with shape $k$ and scale $\theta$, and $d$ is a random variable with distribution $f_d(x)=\frac {2x}{R^2}, 0\le x \le R$.
    To find the CDF of $U_i$ $$F_{U_i}(y)=P(U_i<y)=\int_0^R F_g(yx^\alpha)dx$$ where $F_g(y)$ is the CDF of gamma distributed random variable.
    After performing the integration, which requires the use of the expansion of incomplete gamma function, I came to the following CDF: $$F_{U_i}(y)=\sum_{n=0}^\infty \frac {2(-1)^nR^{k\alpha+n\alpha}y^{k+n}}{\Gamma(k)n!\theta^{k+n}(k+n)(k\alpha+n\alpha+2)}$$ I want the CDF in this format because my ultimate goal is to find the CDF of $U=\sum_{i=1}^NU_i$ which requires me to find the Laplace transform of $F_{U_i}(y)$ and raise it to power $N$.
    Now it's easy to find the PDF of $U_i$, i.e. $f_{U_i}(y)=\frac {d}{dy}F_{U_i}(y)$.
    I'm looking for someone to clarify these points to me:

  • My derived $f_{U_i}(y)$ doesn't make sense to me, because it's based on a divergent series. How can it's integration from zero to infinity equals one?
  • In Matlab I took the $K^{th}$ elements of the series, and plotted it along with the histogram of 1000 $U_i$'s, the two plots are not related.
  • If you look at the definition of $U_i$, $U_i$ can take values from 0 to infinity, high values take less probability. This fact is not clear in the derived distribution $f_{U_i}(y)$. What is the domain of the derived PDF?  
  • How can I simulate the derived PDF in Matlab to check if it's really correlated with $U_i$

    • I'm interested in finding the CDF and PDF of $U_i$ defined as follows, $$U_i=\frac g{d^{\alpha}}$$ where $g$ is a gamma distributed random variable with shape $k$ and scale $\theta$, and $d$ is a random variable with distribution $f_d(x)=\frac {2x}{R^2}, 0\le x \le R$.
    To find the CDF of $U_i$ $$F_{U_i}(y)=P(U_i<y)=\int_0^R F_g(yx^\alpha)dx$$ where $F_g(y)$ is the CDF of gamma distributed random variable.
    After performing the integration, which requires the use of the expansion of incomplete gamma function, I came to the following CDF: $$F_{U_i}(y)=\sum_{n=0}^\infty \frac {2(-1)^nR^{k\alpha+n\alpha}y^{k+n}}{\Gamma(k)n!\theta^{k+n}(k+n)(k\alpha+n\alpha+2)}$$ I want the CDF in this format because my ultimate goal is to find the CDF of $U=\sum_{i=1}^NU_i$ which requires me to find the Laplace transform of $F_{U_i}(y)$ and raise it to power $N$.
    Now it's easy to find the PDF of $U_i$, i.e. $f_{U_i}(y)=\frac {d}{dy}F_{U_i}(y)$.
    I'm looking for someone to clarify these points to me:

  • My derived $f_{U_i}(y)$ doesn't make sense to me, because it's based on a divergent series. How can it's integration from zero to infinity equals one?
  • In Matlab I took the $K^{th}$ elements of the series, and plotted it along with the histogram of 1000 $U_i$'s, the two plots are not related.
  • If you look at the definition of $U_i$, $U_i$ can take values from 0 to infinity, high values take less probability. This fact is not clear in the derived distribution $f_{U_i}(y)$. What is the domain of the derived PDF? What is the integration limit if I wannna find $F_\epsilon(\epsilon)=\int f(y)F_{U_i}(\epsilon(y+\sigma^2))dy$? where $f(y)$ is another density function.
  • How can I simulate the derived PDF in Matlab to check if it's really correlated with $U_i$
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    William
    William
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    William
    William
    • 127
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