MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given the vector $\mathbf{d}$, where $\mathbf{d}\in\mathbb{C}^{N\times 1}$, we have two variables

$X = \mid\mathrm{F}[d]\mid^2,\quad\quad X\ge 0$

$Y = a+b (\mathrm{d}^H\mathrm{d})\quad Y\ge 0$

where :

$\mid.\mid$: denotes the magnitude operator,

$\mathrm{F}[.]$ is the Fourier Transform,

$(.)^H$ is the conjugate transpose operator

both $a$ and $b$ are real constants.

I have got the probability distribution function for $X$, $f_X(x)$ and for $Y$, similarly denoted by $f_Y(y)$. What is the joint probability function F(x,y) and how do I get the Expectation of a function of both X and Y denoted by $g(X,Y)$.

share|cite|improve this question
Parseval's theorem gives that $\langle d,d\rangle=\langle F[d],F[d]\rangle$. – Yoav Kallus Feb 2 '12 at 15:43
yes, Yoav Parseval theorem is right, but the probability distribution function for $\mid[F[d]\mid^2$ is not similar to the distribution of $d^Hd$. – Remy Feb 2 '12 at 16:00
Can you spell out what you mean by $|[F[d]|^2$ then? Do you mean the vector given by element-wise magnitudes? Also, is $Y$ a number or a matrix? If you say $\mathbf{d}$ is $N\times 1$, then $\mathbf{d}^H\mathbf{d}$ is usually interpreted as a number. – Yoav Kallus Feb 2 '12 at 16:19
Yes, sure, happy to clarify. $\mid F[\mathbf{d}]\mid^2$ is the magnitude square of the fourier transform of $d$, i.e, $\mid F[\mathbf{d}]\mid^2=d^H\mathbf{A}d$, where $A$ is a square matrix that mainly contains roots of unity. Therefor $\mid F[\mathbf{d}]\mid^2$ is a scaled sum of squares (scaled by the entries of $A$) and $d^Hd$ is simply a sum of squares. – Remy Feb 2 '12 at 16:29
yes, $\mid F[d]\mid^2$ is a number and $d^Hd$ is a number as well. – Remmy 0 secs ago – Remy Feb 2 '12 at 16:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.