# Confusion about expectation / moment generating function / distributation

(I am currently studying a high dimensional probability course with very little background knowledge in probability theory as a whole, so I hope it's not annoying that I seem to be oblivious about basic concepts, yet am using more involved ideas. Note: I have a good background understanding of measure theory.)

I am having difficulty understanding how to calculate expectation in the following way:

So, by definition I understand that formally $$\mathbb{E}[X]:=\int_\Omega{}X(\omega)d\mathbb{P}(\omega).$$

And that moment generating function is defined as $$M_X(\lambda):=\mathbb{E}[\exp(\lambda{}X)]$$, and this is unique, so if two random variables have the same $$M_X(\lambda)$$ their distributions coincide. Now I am trying to show that the following random variable is normally distributed:

Let $$Y$$ be a random Gaussian vector and $$u\in\mathbb{R}^n$$ (each of its components are standard normally distributed). I am trying to show that $$\langle Y,u\rangle$$~$$N(0,\|u\|_2^2)$$ (where $$\langle\cdot,\cdot\rangle$$ is the standard Euclidean scalar product).

I have shown that the mean is 0 and variance is $$\|u\|_2^2$$ but from my understand this isn't enough. How would I calculate the moment generating function of $$\langle Y,u\rangle$$ and show that this coincides with that of a normal distribution, or is there an easier way of doing so?

So you know that $$\langle Y,u\rangle = \sum_{i=1}^n u_i Y_i$$ (that's the usual inner product over $$\mathbb{R}^n$$), so that for every $$\lambda \in \mathbb{R}$$ $$\mathbb{E}[e^{\lambda \langle Y,u\rangle}] = \mathbb{E}[e^{\lambda \sum_{i=1}^n u_i Y_i}]= \mathbb{E}[\prod_{i=1}^n e^{\lambda u_i Y_i}]$$ Now, you can use the fact that the $$Y_i$$'s are independent (and therefore the $$e^{\lambda u_i Y_i}$$'s are independent) to get $$\mathbb{E}[e^{\lambda \langle Y,u\rangle}] = \prod_{i=1}^n \mathbb{E}[e^{\lambda u_i Y_i}]$$ But each $$Y_i$$ is a standard normal, so you can explicitly compute $$\mathbb{E}[e^{t Y_i}]$$ for any $$t$$ (in particular for $$t=\lambda u_i$$).
Can you take it from there to compute $$\mathbb{E}[e^{\lambda \langle Y,u\rangle}]$$ and compare the resulting expression that that of the MGF of a (univariate) $$\mathcal{N}(0,\lVert u\rVert_2^2)$$?
In more detail: recall that the MGF of $$Z\sim \mathcal{N}(0,\sigma^2)$$ is given by $$\forall \lambda,\; \mathbb{E}[e^{\lambda Z}] = e^{\frac{1}{2}\lambda^2\sigma^2} \tag{\dagger}$$ so that, for every $$1\leq i\leq n$$, since $$Y_i \sim \mathcal{N}(0,1)$$ we have $$\mathbb{E}[e^{\lambda u_i Y_i}] = e^{\frac{1}{2}\lambda^2u_i^2}$$ and therefore $$\mathbb{E}[e^{\lambda \langle Y,u\rangle}] = \prod_{i=1}^n \mathbb{E}[e^{\lambda u_i Y_i}] = \prod_{i=1}^n e^{\frac{1}{2}\lambda^2u_i^2} = e^{\frac{1}{2}\lambda^2 \sum_{i=1}^nu_i^2} = e^{\frac{1}{2}\lambda^2 \lVert u\rVert_2^2} \tag{\ddagger}$$ No, compare $$(\ddagger)$$ to $$(\dagger)$$ to conclude about the distribution of $$\langle Y,u\rangle$$ based on its MGF.
• Let me expand then. Are you familiar with the MGF of a univariate Gaussian (mean 0 and variance $\sigma^2$?) @kam – Clement C. May 11 '20 at 20:06
• Im assuming by multivariate you mean $X$~$N(0,\mathbb{1}_n)$ (with $\mathbb{1}_n$ the identity nxn matrix). I am familiar with that, and the probability density function for it, but not the MGF. – kam May 11 '20 at 20:10