ISI MStat PSB 2009 Problem 4 | Polarized to Normal

This is a very beautiful sample problem from ISI MStat PSB 2009 Problem 4. It is based on the idea of Polar Transformations, but need a good deal of observation o realize that. Give it a Try it !

Problem- ISI MStat PSB 2009 Problem 4

Let $R$ and $\theta$ be independent and non-negative random variables such that $R^2 \sim {\chi_2}^2$ and $\theta \sim U(0,2\pi)$ . Fix $\theta_o \in (0,2\pi)$ . Find the distribution of $R\sin(\theta+\theta_o)$ .

Prerequisites

Convolution

Polar Transformation

Normal Distribution

Solution :

This problem may get nasty, if one try to find the required distribution, by the so-called CDF method. Its better to observe a bit, before moving forward!! Recall how we derive the probability distribution of the sample variance of a sample from a normal population ??

Yes, you are thinking right, we need to use Polar Transformation !!

But, before transforming lets make some modifications, to reduce future complications,

Given, $\theta \sim U(0,2\pi)$ and $\theta_o$ is some fixed number in $(0,2\pi)$ , so, let $Z=\theta+\theta_o \sim U(\theta_o,2\pi +\theta_o)$ .

Hence, we need to find the distribution of $R\sin Z$ . Now, from the given and modified information the joint pdf of $R^2$ and $Z$ are,

$f_{R^2,Z}(r,z)=\frac{r}{2\pi}exp(-\frac{r^2}{2}) \ \ R>0, \theta_o \le z \le 2\pi +\theta_o$

Now, let the transformation be $(R,Z) \to (X,Y)$ ,

$X=R\cos Z \\ Y=R\sin Z$ , Also, here $X,Y \in \mathbb{R}$

Hence, $R^2=X^2+Y^2 \\ Z= \tan^{-1} (\frac{Y}{X})$

Hence, verify the Jacobian of the transformation $J(\frac{r,z}{x,y})=\frac{1}{r}$ .

Hence, the joint pdf of $X$ and $Y$ is,

$f_{X,Y}(xy)=f_{R,Z}(x^2+y^2, \tan^{-1}(\frac{y}{x})) J(\frac{r,z}{x,y}) \\ =\frac{1}{2\pi}exp(-\frac{x^2+y^2}{2})$ , $x,y \in \mathbb{R}$ .

Yeah, Now it is looking familiar right !!

Since, we need the distribution of $Y=R\sin Z=R\sin(\theta+\theta_o)$ , we integrate $f_{X,Y}$ w.r.t to $X$ over the real line, and we will end up with, the conclusion that,

$R\sin(\theta+\theta_o) \sim N(0,1)$ . Hence, We are done !!

Food For Thought

From the above solution, the distribution of $R\cos(\theta+\theta_o)$ is also determinable right !! Can you go further investigating the occurrence pattern of $\tan(\theta+\theta_o)$ ?? $R$ and $\theta$ are the same variables as defined in the question.

Give it a try !!

Similar Problems and Solutions

ISI MStat PSB 2008 Problem 10 — Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Related

This is a very beautiful sample problem from ISI MStat PSB 2009 Problem 4. It is based on the idea of Polar Transformations, but need a good deal of observation o realize that. Give it a Try it !

Problem- ISI MStat PSB 2009 Problem 4

Let $R$ and $\theta$ be independent and non-negative random variables such that $R^2 \sim {\chi_2}^2$ and $\theta \sim U(0,2\pi)$ . Fix $\theta_o \in (0,2\pi)$ . Find the distribution of $R\sin(\theta+\theta_o)$ .

Prerequisites

Convolution

Polar Transformation

Normal Distribution

Solution :

This problem may get nasty, if one try to find the required distribution, by the so-called CDF method. Its better to observe a bit, before moving forward!! Recall how we derive the probability distribution of the sample variance of a sample from a normal population ??

Yes, you are thinking right, we need to use Polar Transformation !!

But, before transforming lets make some modifications, to reduce future complications,

Given, $\theta \sim U(0,2\pi)$ and $\theta_o$ is some fixed number in $(0,2\pi)$ , so, let $Z=\theta+\theta_o \sim U(\theta_o,2\pi +\theta_o)$ .

Hence, we need to find the distribution of $R\sin Z$ . Now, from the given and modified information the joint pdf of $R^2$ and $Z$ are,

$f_{R^2,Z}(r,z)=\frac{r}{2\pi}exp(-\frac{r^2}{2}) \ \ R>0, \theta_o \le z \le 2\pi +\theta_o$

Now, let the transformation be $(R,Z) \to (X,Y)$ ,

$X=R\cos Z \\ Y=R\sin Z$ , Also, here $X,Y \in \mathbb{R}$

Hence, $R^2=X^2+Y^2 \\ Z= \tan^{-1} (\frac{Y}{X})$

Hence, verify the Jacobian of the transformation $J(\frac{r,z}{x,y})=\frac{1}{r}$ .

Hence, the joint pdf of $X$ and $Y$ is,

$f_{X,Y}(xy)=f_{R,Z}(x^2+y^2, \tan^{-1}(\frac{y}{x})) J(\frac{r,z}{x,y}) \\ =\frac{1}{2\pi}exp(-\frac{x^2+y^2}{2})$ , $x,y \in \mathbb{R}$ .

Yeah, Now it is looking familiar right !!

Since, we need the distribution of $Y=R\sin Z=R\sin(\theta+\theta_o)$ , we integrate $f_{X,Y}$ w.r.t to $X$ over the real line, and we will end up with, the conclusion that,

$R\sin(\theta+\theta_o) \sim N(0,1)$ . Hence, We are done !!

Food For Thought

From the above solution, the distribution of $R\cos(\theta+\theta_o)$ is also determinable right !! Can you go further investigating the occurrence pattern of $\tan(\theta+\theta_o)$ ?? $R$ and $\theta$ are the same variables as defined in the question.

Give it a try !!

Similar Problems and Solutions

ISI MStat PSB 2008 Problem 10 — Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Related

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