$$X = \frac1N \sum_1^N d_i \cos\theta_i $$ $$Y = \frac1N \sum_1^N d_i \sin\theta_i $$
$$d_i \sim LN(m_i, \sigma^2) $$ (LN mean Log normal distribution, $m_i$ can be measured, actually function of measured value)
$$\theta_i \sim U(-\pi, \pi) $$ (U mean uniform distribution)
PDF of $cos\theta_i$, $$f_C(c)=\frac{1}{\pi\sqrt{1-c^2}} c \in [-1, 1] $$
$d_i$ and $cos\theta_i$ are independent.
If $N$ is large enough, Random variables $X$ and $Y$ are approximated as Gaussian Random variables
What is the PDF of $\sin\theta_i$, $f_S(s)$ is same form of $f_C(c)$?
What is the covariance or correlation coefficient of $X$ and $Y$ ?
(For the first Question, I think It has same form but I can't show this) (For the second Question, $$E[(X-m_x)(Y-m_y)]=\frac{1}{N^2}E[\sum_1^N d_i\cos\theta_i\sum_1^N d_j\sin\theta_j]$$ $$ E[d_i\cos\theta_id_j\sin\theta_j]=E[d_i^2]E[\cos\theta_i\sin\theta_i], i=j $$ $$ E[\cos\theta_i\sin\theta_i]=\frac 12 E[\sin2\theta_i]=0 $$ Is this really $E[\cos\theta_i\sin\theta_i]=0$? If it is true, I mean $E[\cos\theta_i\sin\theta_i]=0$, then $E[(X-m_x)(Y-m_y)]=0$ That means $X$ and $Y$ are independent. I don't understand this result. Because $X$ and $Y$ have 2 same random variable.
I am so stuck in this problem Anyone can save my life? I really appreciate you even read this quite so long questions and think about this problem. Have a nice one everyone !
