Hi,

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

Thanks a ton

Regards,

Results 1 to 8 of 8

- Apr 3rd 2009, 09:39 AM #1

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- Apr 3rd 2009, 10:04 AM #2
Hello,

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

- Apr 3rd 2009, 10:16 AM #3

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- Apr 3rd 2009, 10:29 AM #4
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 ?

- Apr 3rd 2009, 11:52 AM #5

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- Apr 4th 2009, 08:44 AM #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, $\displaystyle \mathbb{E}(h(X))=\int f(x)h(x) ~dx$

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 : $\displaystyle \frac{1}{\sqrt{2\pi}} \cdot e^{-x^2/(2\sigma^2)} \times \frac{1}{\sqrt{2 \pi}} \cdot e^{-y^2/(2\sigma^2)}=\frac{1}{2\pi} \exp\left(-\frac{x^2+y^2}{2\sigma^2}\right)$

$\displaystyle

\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 :

$\displaystyle \phi ~:~ (x,y) \mapsto (u,v)=\left(x^2+y^2,\tfrac{x}{\sqrt{x^2+y^2}}\righ t)$

$\displaystyle \phi ~:~ \mathbb{R} \times \mathbb{R}-\{(0,0)\} \to \mathbb{R}_+ \times \mathbb{R}$

(but actually you don't have to care about the (0,0) point)

The Jacobian matrix is :

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

The Jacobian of the transformation is thus :

$\displaystyle |\det J_\phi|=\left|2x \cdot \frac{-xy}{(x^2+y^2)^{3/2}}-2y \cdot \frac{y^2}{(x^2+y^2)^{3/2}}\right|=2 |y| \left|\frac{x^2+y^2}{(x^2+y^2)^{3/2}}\right|=2 |y| \cdot \frac{1}{\sqrt{v}}$

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

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

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

And the Jacobian is finally :

$\displaystyle \boxed{2 \cdot \sqrt{u-uv^2} \cdot \frac{1}{\sqrt{v}}}$

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.

- Apr 4th 2009, 10:39 AM #7

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- Apr 5th 2009, 12:36 AM #8
And so what's wrong ? Can't you finish it ?

The pdf is :

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

Since $\displaystyle u=x^2+y^2$, this is :

$\displaystyle \frac{1}{2\pi} \exp\left(-\frac{u}{2\sigma^2}\right)$

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

So the pdf of (U,V) is :

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

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