Suppose $x_i$ are independent standard Gaussian random variables with mean 0 and variance 1. Moreover set $\lambda_i>0$ and $t$ constant. Compute $\mathbb{E}[e^{t |\sum_{i=1}^n \lambda_ix_i|}]$.
I know that the distribution function of a standard Gaussian random variable is given by $f(x_i)=\frac{1}{\sqrt{2\pi}}e^{-x_i^2/2}$. I started computing the integral $\int_{- \infty}^{\infty} e^{t | \sum_{i=1}^n \lambda_i x_i|} f(x_i) dx$, but I am having the following problem. Since we are not considering $\sum{\lambda_i x_i}$ but $|\sum \lambda_i x_i|$ I cannot split the integral into single ones regarding the independent $x_i$. Hence, I am confused on how to exactly compute the integral.
May someone please help me?
Hint: First, show that $Z = \displaystyle\sum_{i = 1}^{n}\lambda_ix_i$ is a normal random variable with mean $0$ and variance $\sigma^2 = \displaystyle\sum_{i = 1}^{n}\lambda_i^2$.
Once you've done this, the PDF of $Z$ is $f(z) = \dfrac{1}{\sigma\sqrt{2\pi}}e^{-z^2/(2\sigma^2)}$, and you need to calculate $$\mathbb{E}[e^{t|Z|}] = \int_{-\infty}^{\infty}e^{t|z|}f(z)\,dz = 2\int_{0}^{\infty}e^{tz} \cdot \dfrac{1}{\sigma\sqrt{2\pi}}e^{-z^2/(2\sigma^2)}\,dz.$$