# Convolution between normal distribution and the maximum over $m$ Gaussian draws

$$\DeclareMathOperator\erf{erf}$$ Let's consider the Gaussian distribution $$P_X(x)= \frac{1}{\sqrt{2 \pi \sigma^2}} e^{- \frac{x^2}{2 \sigma^2}}$$. Now consider the random variable $$W \equiv \max \{ X_1, \dots , X_m \}$$, where $$X_i \sim P_X(x)$$. So $$w$$ indicates the maximum over $$m$$ i.i.d. Gaussian draws; we call $$P_W(w)$$ the corresponding distribution. $$P_W(w)$$ can be easily calculated from the formula $$P_Y(y)=mP_X(y)F_X(y)^{m-1}$$, where $$F_X(y)$$ is the c.d.f. of $$P_X(x)$$. We get:

$$P_Y(y)= m \frac{1}{\sqrt{2 \pi \sigma^2}} e^{- \frac{y^2}{2 \sigma^2}} \left(\frac{1}{2} \left( 1 + \erf \left( \frac{y}{\sqrt{2} \sigma}\right) \right) \right)^{m -1}$$

I'm interested in the convolution between $$P_X(x)$$ and $$P_W(w)$$, i.e.:

$$I(\Delta) = \int_{-\infty}^\infty P_X(x') P_W(x'+ \Delta) \, dx'$$

Is it possible to get an analytical closed form or even an approximated solution for $$I(\Delta)$$?

• It seems odd that you initially call the random variable $w=\max\{x_1,\ldots,x_m\}$ but then write $P_W(w),$ as if (capital) $W$ is the random variable and (lower-case) $w$ is the argument to its density function. Dec 24, 2022 at 9:12
• Do you have a response to the answer below? Dec 24, 2022 at 22:24
• I only read the headline, but won't the convolution with the max give the distribution of X + Max, and isn't that the distribution of max(X+X_1, ..., X + X_n) ?
– mike
Dec 31, 2022 at 9:45

Mathematica cannot find such an expression even when $$\sigma=1$$, $$m=3$$, and $$\Delta=0$$: