If $\mathbb E[X\mid Y]$ can be seen as a projection why $\mathbb E[X\mid Y]=\frac{\mathbb E[XY]}{\mathbb E[Y^2]}Y$ is not always true?

Question

We know that $\mathbb E[XY]$ is a scalaire product on $L^2(\mathbb P)$. In a book (an introduction to stochastic differential equation of Evans) page 30-31, it's written that $\mathbb E[X\mid Y]$ can be seen as the orthogonal projection of $X$ on $Y$. So, why $$\mathbb E[X\mid Y]=\frac{\mathbb E[XY]}{\mathbb E[Y^2]}Y,$$ not always true ?

Answer 1

You are mixing up the notions of projection on the one-dimensional vector space spanned by $Y$ and projection on the space of all random variables measurable with respect to $\sigma (Y)$.

Answer 2

I'm sure that in your book it's written that "$\mathbb E[X\mid Y]$ it the orthogonal projection onto $\sigma (Y)$" instead of "on $Y$". If $V$ is a finite vector, $W$ a subspace of $V$ and $\{w_1,...,w_n\}$ an orthogonal basis, then indeed $$Proj_W(v)=\sum_{i=1}^n\frac{\left<v,w_i\right>}{\|w_i\|^2}w_i.$$

This can be prolonged on $V$ and $W$ with infinite dimension with certain conditions (like Hilbert). Now, what would be an orthogonal basis of $L^2(\mathbb P,\sigma (Y))$ ?

Answer 3

Well if you take $X$ and $Y$ to be independent, the equation reads $$1 = \frac{Y\mathbb{E}(Y)}{\mathbb{E}(Y^2)},$$ which seems a little bit weird.