Hi,

I am clueless on how to do the attached questions...Will be grateful if someone could help me here...

Thanks a ton

Regards,

2. Hello,
Originally Posted by nodi1310
Hi,

I am clueless on how to do the attached questions...Will be grateful if someone could help me here...

Thanks a ton

Regards,
Question 1) is just the law of large numbers : Proof of the law of large numbers - Wikipedia, the free encyclopedia

3. Hi,

For Qn 1) Could u pls b a more descriptive?? As i am not getting the proof using that..

Thanks alot

Regards,

4. Originally Posted by nodi1310
Hi,

For Qn 1) Could u pls b a more descriptive?? As i am not getting the proof using that..

Thanks alot

Regards,
This is quite surprising that you don't understand the proof since it's just Chebyshev's inequality...

You can just say that by the law of large numbers, the probability goes to 1.
It's the exact result of the LLN (provided that all conditions are satisfied)

Or you can directly apply Chebyshev's inequality.

It depends on what you have been taught.

For question 2) : which method have you been taught for finding the joint pdf of two variables ?

5. Hi,

I think i figured out Qn 1) problem is they havent taught Chebychev inequality yet,they teach next week..tats y i got confused...

For Qn 2) We were taught the method where we use the double integrals and partial derivatives.

Thanks alot

Regards,

6. For question 2), this is how I've been taught. It's similar to yours, according to what you have said, but it's difficult for me to adapt it...

There is a formula that says :
X has a pdf f iff for any bounded and continuous function h, $\mathbb{E}(h(X))=\int f(x)h(x) ~dx$
Same goes for the density of a couple.

So here, we'll find the density of (U,V) :

Let h be a bounded and continuous function.
Note that since X and Y are independent, the pdf of (X,Y) is the product of the densities : $\frac{1}{\sqrt{2\pi}} \cdot e^{-x^2/(2\sigma^2)} \times \frac{1}{\sqrt{2 \pi}} \cdot e^{-y^2/(2\sigma^2)}=\frac{1}{2\pi} \exp\left(-\frac{x^2+y^2}{2\sigma^2}\right)$


\begin{aligned} \mathbb{E}(h(U,V))
&=\mathbb{E}\left(h\left(X^2+Y^2,\tfrac{X}{\sqrt{X ^2+Y^2}}\right)\right) \\
&=\iint_{\mathbb{R}^2} h\left(x^2+y^2,\tfrac{x}{\sqrt{x^2+y^2}}\right) \cdot \frac{1}{2\pi} \exp\left(-\tfrac{x^2+y^2}{2\sigma^2}\right) ~dx dy
\end{aligned}

Now let the transformation :
$\phi ~:~ (x,y) \mapsto (u,v)=\left(x^2+y^2,\tfrac{x}{\sqrt{x^2+y^2}}\righ t)$
$\phi ~:~ \mathbb{R} \times \mathbb{R}-\{(0,0)\} \to \mathbb{R}_+ \times \mathbb{R}$
(but actually you don't have to care about the (0,0) point)

The Jacobian matrix is :

$J_\phi=\begin{pmatrix} \frac{\partial u}{\partial x} & & \frac{\partial u}{\partial y} \\ & & \\ \frac{\partial v}{\partial x} & & \frac{\partial v}{\partial y} \end{pmatrix}=\dots=\begin{pmatrix} 2x & 2y \\ \frac{y^2}{(x^2+y^2)^{3/2}} & \frac{-xy}{(x^2+y^2)^{3/2}}\end{pmatrix}$

The Jacobian of the transformation is thus :
$|\det J_\phi|=\left|2x \cdot \frac{-xy}{(x^2+y^2)^{3/2}}-2y \cdot \frac{y^2}{(x^2+y^2)^{3/2}}\right|=2 |y| \left|\frac{x^2+y^2}{(x^2+y^2)^{3/2}}\right|=2 |y| \cdot \frac{1}{\sqrt{v}}$

But $x^2+y^2=u \Rightarrow |y|=\sqrt{u-x^2}$

And $v=\frac{x}{\sqrt{x^2+y^2}}=\frac{x}{\sqrt{u}} \Rightarrow x^2=uv^2$

Hence $|y|=\sqrt{u-uv^2}$

And the Jacobian is finally :
$\boxed{2 \cdot \sqrt{u-uv^2} \cdot \frac{1}{\sqrt{v}}}$

I'm sure you need the Jacobian for your method... Can you actually finish it ? From here, it all depends on your "formula". So show me and I'll help you.

7. Hi,

Thank u very much...I have attached the formula tat v use...Please help

Regards,

8. And so what's wrong ? Can't you finish it ?

The pdf is :

$\frac{1}{2\pi} \exp\left(-\frac{x^2+y^2}{2\sigma^2}\right)$

Since $u=x^2+y^2$, this is :
$\frac{1}{2\pi} \exp\left(-\frac{u}{2\sigma^2}\right)$

And the Jacobian part is the inverse of $|\det J_\phi|$, which is $\frac{\sqrt{v}}{2 \cdot \sqrt{u} \cdot \sqrt{1-v^2}}$

So the pdf of (U,V) is :

$\frac{1}{2\pi} \exp\left(-\frac{u}{2\sigma^2}\right) \cdot \frac{\sqrt{v}}{2 \cdot \sqrt{u} \cdot \sqrt{1-v^2}}=\frac{1}{2\pi} \left(\frac{\exp\left(-\frac{u}{2\sigma^2}\right)}{\sqrt{u}}\right) \cdot \left(\frac{\sqrt{v}}{2 \sqrt{1-v^2}}\right)$

which is a product of 2 functions, each depending on u or v, separately. This proves that U and V are independent.