# Approximate sum of gamma distributions using normal distribution

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.

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.
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
• @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. – Felix Marin Jan 27 '17 at 1:28