$p$-norm of random variables and weighted $L^p$ space resemblance

Question

I noticed a very similar relationship between weighted $L^p$ space (denoted $L_w^p$) and normed vector space of random variables. I want to unify these two spaces but there always seems to be a difference, or are they actually different?

Recall that $L_w^p$-norm is defined if the following integral exists: $$ \left\| f \right\|_{L_w^p} := \left( \int |f(x)|^p w(x) \, \mathrm{d}x \right)^{\frac{1}{p}}. $$ The $p$-norm of a real-valued R.V. $X: \Omega \to \mathbb{R}$ is defined to be: $$ \left\| X \right\|_p := \mathbb{E}[|X|^p]^{\frac{1}{p}} = \left(\int_\Omega |X|^p \, \mathrm{d}\mathbb{P}\right)^{\frac{1}{p}} = \left(\int_\mathbb{R} |x|^p f_X(x) \, \mathrm{d}x\right)^{\frac{1}{p}} \label{1}\tag{1} $$ where $(\Omega, \mathscr{A}, \mathbb{P})$ is a probability space and $f_X(x)$ is the p.d.f. of R.V. $X$.

You can prove that these two norms are indeed norms with triangle inequality: $$ \left\| f+g \right\|_{L_w^p} \le \left\| f \right\|_{L_w^p} + \left\| g \right\|_{L_w^p} \label{2}\tag{2} $$ and $$ \left\| X+Y \right\|_p \le \left\| X \right\|_p + \left\| Y \right\|_p. \label{3}\tag{3} $$ Note that $f_X(x)$ in equation \eqref{1} can be viewed as the "weight" of $|X|^p$. Therefore, (using weighted norm in $L_w^p$ space), the triangle inequality \eqref{3} can be written as: $$ \left\| Z \right\|_{L_{h_Z(z)}^p} \le \left\| X \right\|_{L_{f_X(x)}^p} + \left\| Y \right\|_{L_{g_Y(y)}^p} $$ where $Z=X+Y$ is another R.V., $f_X(x)$, $g_Y(y)$ and $h_Z(z)$ is the p.d.f. of R.V. $X$, $Y$ and $Z$ respectively.

However, this is NOT the triangle inequality in $L_w^p$ space since the norm operation was done in three different spaces: $L_{h_Z(z)}^p$, $L_{f_X(x)}^p$, $L_{g_Y(y)}^p$. But it really resembles equation \eqref{2}. How could I understand this?

Answer 1

I don't know what's so surprising about this, but maybe I misunderstood the question. Both inequalities are just instances of the general triangle inequality on $L^p(\Omega, \mathcal{F}, \mu)$ for some measure space $(\Omega, \mathcal{F}, \mu)$, see https://en.wikipedia.org/wiki/Minkowski_inequality.

In the first case the corresponding measure space is e.g. $\mathbb{R}$ equipped with the Borel or Lebesgue $\sigma$-algebra and measure $\mu(\mathrm{d}x) = w(x) \mathrm{d}x$.

In the second case the corresponding measure space is the probability space $(\Omega, \mathcal{F}, \mathbb{P})$ on which $X, Y, Z$ are defined.

Of course it is not true that every random variable has a density w.r.t. Lebesgue measure as you implicitly assumed and even if $X, Y$ possess a density, $Z = X+Y$ doesn't have to (think $Y = -X$ for example).

Even assuming this, it is not correct that \eqref{3} can be written as $$ \left\| Z \right\|_{L_{h_Z(z)}^p} \le \left\| X \right\|_{L_{f_X(x)}^p} + \left\| Y \right\|_{L_{g_Y(y)}^p}. $$ First, this does not even make sense, because in general $\Omega \neq \mathbb{R}$. The correct version would be (assuming the densities exist) $$ \left(\int_{\mathbb{R}} \vert x \vert^p f_Z(x) \,\mathrm{d}x\right)^{1/p} = \Vert X + Y\Vert_{L^p(\Omega, \mathcal{F}, \mu)} \leq \Vert X \Vert_{L^p(\Omega, \mathcal{F}, \mu)} + \Vert Y\Vert_{L^p(\Omega, \mathcal{F}, \mu)} \\ = \left(\int_{\mathbb{R}} \vert x \vert^p f_X(x) \,\mathrm{d}x\right)^{1/p} + \left(\int_{\mathbb{R}} \vert x \vert^p f_Y(x) \,\mathrm{d}x\right)^{1/p}, $$ which is evidently not in the form of (2).

If $X, Y$ have a joint density $f_{X, Y}$, then $$ \Vert X + Y\Vert_{L^p(\Omega, \mathcal{F}, \mu)} = \mathbb{E}[\vert X + Y\vert^p]^{1/p} = \left(\int_{\mathbb{R}^2} \, \vert x+y\vert^p f_{X, Y}(x, y) \,\mathrm{d}x\,\mathrm{d}y\right)^{1/p} \leq \left(\int_{\mathbb{R}^2} \, \vert x\vert^p f_{X, Y}(x, y) \,\mathrm{d}x\,\mathrm{d}y\right)^{1/p} + \left(\int_{\mathbb{R}^2} \, \vert y\vert^p f_{X}(x, y) \,\mathrm{d}x\,\mathrm{d}y\right)^{1/p}\\ = \left(\int_{\mathbb{R}} \, \vert x\vert^p f_{X}(x) \,\mathrm{d}x\,\mathrm{d}y\right)^{1/p} + \left(\int_{\mathbb{R}} \, \vert y\vert^p f_{Y}(y) \,\mathrm{d}x\,\mathrm{d}y\right)^{1/p}\\ = \Vert X \Vert_{L^p(\Omega, \mathcal{F}, \mu)} + \Vert Y\Vert_{L^p(\Omega, \mathcal{F}, \mu)}, $$ hence you can understand at least this case using the triangle inequality for a weighted $L^p$ space on $\mathbb{R}^2$ instead of $\mathbb{R}$ (and the general case could then be recovered by extending the probability space and adding an independent Gaussian $Z \sim \mathcal{N}(0, \epsilon I_2)$ to $(X, Y)$).

Answer 2

In what you're calling the $L_w^p$-norm, I assume $w:\mathbb R\to \mathbb R_+$ is some measurable function, and the integral defining $\Vert\cdot\Vert_{L_w^p}$ is the Lebesgue integral.

If this is what you meant, then these are both special cases of an $L^p$ norm on a measure space. Given a measure space $(M,\Sigma,\mu)$, the space $L^p(M,\Sigma,\mu)$ is the space of $\Sigma$-measurable functions $f:M\to\mathbb R$, modulo $\mu$-almost everywhere equivalence, for which the Lebesgue integral $$\left\Vert \int |f|^p \text{ d}\mu\right\Vert$$ is finite.

Your $L_w^p$-norm is the special case where the underlying measurable space is the reals, and the measure is given by $\mu(A):=\int_A w(x) \text{ d}x$. The random variable one is just what I've written here for the special case that $\mu$ is a probability measure.

[I phrased things for $1\leq p <\infty$, but analogous statements hold for $p=\infty$ too.]