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If $X_1$, $X_2$, $X_3$, and $X_4$ are gamma distributions with $\theta=2$ and $\alpha_1=3, \alpha_2=2, \alpha_3=5, \alpha_4=3$, respectively, and $Y=X_1+X_2+X_3+X_4$, I can find P(Y$\leq$25) by integrating the gamma function Y, and I obtain 0.4810.

However, I am asked to use a normal distribution to approximate the answer.

I cannot figure out how I would use the Central Limit Theorem to do this. Are $X_1$, $X_2$, $X_3$, and $X_4$ really a random sample of size 4 from a distribution with mean 26 and variance 52? My attempt of P($\dfrac{X_{bar}-\dfrac {26}{4}}{\sqrt{52/4}/2}\leq {\dfrac{\dfrac{25}{4}-\dfrac {26}{4}}{\sqrt{52/4}/2}}$) does not work.

Also, I do not know how I would justify using a normal approximation for the gamma distribution.

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I assume you are using the shape-scale parametrization $$f_X(x) = \frac{x^{\alpha-1} e^{-x/\theta}}{\theta^\alpha \Gamma(\alpha)}, \quad x > 0$$ and that $\theta = 2$ is the common scale parameter. In such a case, $Y = X_1 + X_2 + X_3 + X_4$ would be gamma with shape $\alpha_1 + \alpha_2 + \alpha_3 + \alpha_4 = 13$. When $\alpha \gg \theta$, we can use a normal approximation with $\mu = \alpha \theta$, $\sigma^2 = \alpha \theta^2$, to get $$\Pr[Y \le 25] \approx \Pr\left[\frac{Y - \mu}{\sigma} \le \frac{25 - 26}{2\sqrt{13}}\right] \approx \Phi(-0.138675) \approx 0.444853.$$ Is it a good approximation? No, not really.

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The answer is given by

\begin{align} \mc{P} & \equiv \int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty}{x_{1}^{2}\expo{-x_{1}/2} \over 2^{3}\Gamma\pars{3}}\, {x_{2}\expo{-x_{2}/2} \over 2^{2}\Gamma\pars{2}}\, {x_{3}^{4}\expo{-x_{3}/2} \over 2^{5}\Gamma\pars{5}}\,{x_{2}^{2}\expo{-x_{2}/2} \over 2^{3}\Gamma\pars{3}}\times \\[5mm] &\ \bracks{x_{1} + x_{2} + x_{3} + x_{4} \leq 25} \dd x_{1}\,\dd x_{2}\,\dd x_{3}\,\dd x_{4} \\[1cm] & = {1 \over 2^{13}\,\Gamma^{2}\pars{3}\Gamma\pars{2}\Gamma\pars{5}} \int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty} x_{1}^{2}x_{2}x_{3}^{4}x_{2}^{2}\expo{-\pars{x_{1} + x_{2} + x_{3} + x_{4}}/2} \\[5mm] & \times \bracks{x_{1} + x_{2} + x_{3} + x_{4} \leq 25} \dd x_{1}\,\dd x_{2}\,\dd x_{3}\,\dd x_{4} \\[1cm] & = {1 \over 2^{13}\,\Gamma^{2}\pars{3}\Gamma\pars{2}\Gamma\pars{5}} \int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty} x_{1}^{2}x_{2}x_{3}^{4}x_{2}^{2}\expo{-\pars{x_{1} + x_{2} + x_{3} + x_{4}}/2} \times \\[5mm] &\ \bracks{\int_{c - \infty\ic}^{c + \infty\ic} {\exp\pars{\bracks{25 - x_{1} - x_{2} - x_{3} - x_{4}}s} \over s}\, {\dd s \over 2\pi\ic}} \dd x_{1}\,\dd x_{2}\,\dd x_{3}\,\dd x_{4} \\[1cm] & = {1 \over 2^{13}\,\Gamma^{2}\pars{3}\Gamma\pars{2}\Gamma\pars{5}} \int_{c - \infty\ic}^{c + \infty\ic} \bracks{\int_{0}^{\infty}x_{1}^{2}\expo{-\pars{1/2 + s}x_{1}}\,\dd x_{1}}^{2} \bracks{\int_{0}^{\infty}x_{2}\expo{-\pars{1/2 + s}x_{2}}\,\dd x_{2}} \\[5mm] &\ \times\bracks{\int_{0}^{\infty}x_{3}^{4}\expo{-\pars{1/2 + s}x_{3}}\,\dd x_{3}}{\expo{25s} \over s}\,{\dd s \over 2\pi\ic} \\[1cm] & = {1 \over 2^{13}\,\Gamma^{2}\pars{3}\Gamma\pars{2}\Gamma\pars{5}} \int_{c - \infty\ic}^{c + \infty\ic} \bracks{\Gamma\pars{3} \over \pars{1/2 + s}^{3}}^{2} {\Gamma\pars{2} \over \pars{1/2 + s}^{2}} {\Gamma\pars{5} \over \pars{1/2 + s}^{5}}\,{\expo{25 s} \over s} \,{\dd s \over 2\pi\ic} \\[5mm] & = {1 \over 2^{13}}\int_{c - \infty\ic}^{c + \infty\ic} {\expo{25s} \over s\pars{s + 1/2}^{13}}\,{\dd s \over 2\pi\ic} = 1 + {1 \over 2^{13}}\,{1 \over 12!}\, \left.\totald[12]{}{s}\pars{\expo{25s} \over s}\right\vert_{\ s\ =\ -1/2} \\[5mm] & = \bbx{\ds{1 - {10929103729019569 \over 78479622144}\,\expo{-25/2}}} \approx 0.481024781120949\ldots \end{align}

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  • $\begingroup$ Such a lengthy computation is rendered unnecessary by simply recognizing that a gamma distribution with common scale/rate and positive integer shape parameter is the sum of IID exponential distributions; consequently, the sum of independent gamma distributions with common scale/rate is itself gamma, a fact that is also proved quite readily in the noninteger shape case using the moment-generating function. The exact calculation was not what was in question here; rather, the question pertains to the use of a normal approximation to the gamma and its appropriateness. $\endgroup$ – heropup Jan 26 '17 at 23:25
  • $\begingroup$ @heropup I guess ${\large four\ variables}$ isn't enough to claim a good approximation via Central Limit Theorem. Anyway, it's fine to know for newbies how to accomplish an exact calculation. Thanks for your advice. $\endgroup$ – Felix Marin Jan 27 '17 at 1:28

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