# Integrating the pdf of a normal distribution

I need to find the distribution of $Y=X_1+X_2$ where both $X_1$ and $X_2$ are normally distributed with $(\mu,\sigma^2)$.

So I'm looking for $f_y(y)=P(Y=y)$$=P(X_1+X_2=Y)$$=P(X_1=x_1)P(X_2=Y-x_1)$$=\int_{x=-\infty}^{y} \frac1{\sqrt{2\pi}\sigma} e^{\frac{(x_1-\mu)^2}{2\sigma^2}} \frac1{\sqrt{2\pi}\sigma} e^{\frac{(y-x_1-\mu)^2}{2\sigma^2}} dx$$ $$=\int_{x=-\infty}^{y} \frac1{2\pi\sigma^2} e^{\frac{(x_1-\mu)^2+(y-x_1-\mu)^2}{2\sigma^2}} dx$$$$= \frac1{2\pi\sigma^2} \int_{x=-\infty}^{y} e^{\frac{(x_1-\mu)^2+(y-x_1-\mu)^2}{2\sigma^2}} dx$$ which is about where I get stuck. I tried expanding the brackets and removing any terms from the intergral that don't depend on x and what I get is: $$= \frac1{2\pi\sigma^2} e^{\frac{y^2+2\mu^2-2y\mu}{2\sigma^2}}\int_{x=-\infty}^{y} e^{\frac{x_1(x_1-y)}{\sigma^2}} dx$$ I've been looking for a way to use u substitution on the exponent term$\frac{x_1(x_1-y)}{\sigma^2}$but I can't come up with anything that works. Am I missing something obvious? Any nudges in the right direction are welcomed. - Be aware that$f_Y(y)$does not equal$P(Y=y)$and it does not equal$P(X_1=x_1)P(X_2=Y-x_1)$either. Your integral is not correct either: the limits are incorrect as is the integrand. – Dilip Sarwate Jan 30 '13 at 3:59 ## 3 Answers Following up on Robert Israel's suggestion to use$X_i = \mu + \sigma Z_i$where the$Z_i$are independent standard normal random variables, note that$Y = X_1+X_2 = 2\mu + \sigma (Z_1+Z_2)$. So you could just find the density of$Z_1+Z_2$and then transform it into the density of$X_1+X_2$using the standard result for linear transformations, viz. $$f_{a+bW}(w) = \frac{1}{|b|}f_W\left(\frac{w - a}{b}\right)$$ which gives $$f_{X_1+X_2}(y) = f_{Y}(y) = \frac{1}{\sigma}f_{Z_1+Z_2}\left(\frac{y-2\mu}{\sigma}\right).$$ This will save you a lot of carrying around$\mu$and$\sigma$which merely clutter up the formulas. A calculation of the density of$\alpha Z_1 + \beta Z_2$for independent standard normal random variables can be found in the answers to this earlier question. Here, of course, you have$\alpha = \beta = 1$. Note that convolution is not explicitly required to obtain the answer in this simple case; an appeal to the circular symmetry of the joint density of$Z_1$and$Z_2$allows one to deduce the answer easily. - Have you learned about the characteristic function of a random variable? The characteristic function of a sum of independent random variables, in general, is the product of the characteristic functions of the components of the sum. So if we call the characteristic function of a variable$X$as$\Phi_X$, then$\Phi_{X+Y} = \Phi_X \Phi_Y$. The characteristic function of a Normal random variable$X$with mean$\mu_X$and variance$\sigma_X$is $$\Phi_X(\nu) = exp\left({j \nu \mu_X - \frac{\sigma_X^2 \nu^2}{2}}\right)$$ where$j = \sqrt{-1}$. So the sum of two Normal random variables,$X$as above and$Y$with mean$\mu_Y$and variance$\sigma_Y$is $$\Phi_{X+Y}(\nu) = exp\left({j \nu \mu_X - \frac{\sigma_X^2 \nu^2}{2}}\right)exp\left({j \nu \mu_Y - \frac{\sigma_Y^2 \nu^2}{2}}\right).$$ From there you can find your answer. - The sum of independent (a necessary assumption you left out) normal random variables has a normal distribution. Well, maybe the purpose of the exercise is to prove that. It makes things simpler if you first standardize your random variables:$X_i = \mu + \sigma Z_i$where$Z_i$have standard normal distribution (mean$0$and variance$1$). Your line$f_y(y)=P(Y=y)=P(X_1+X_2=Y)=P(X_1=x_1)P(X_2=Y−x_1)\$ is all wrong. These are continuous random variables, and you're looking for a density, not a probability. The correct statement is

$$f_Y(y) = \int_{-\infty}^\infty f_{X_1}(x) f_{X_2}(y-x)\ dx$$

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