# Thread: Expressing a linear function in terms of independent random variables

1. ## Expressing a linear function in terms of independent random variables

Let Y1<Y2<...<Yn be the order statistics of a random sample of size n from the pdf $f(x) = e^{-x}$ x ranging from 0 to infinity.

Demonstrate that all linear functions of Y1, Y2,...,Yn such as $\Sigma a_i Y_i$ can be expressed as a linear function of independent random variables.

so:

$\Sigma a_i Y_i = a_1Y_1 + a_2Y_2 + ... + a_nY_n = a_1e^{-x_1} + a_2e^{-x_2} +...+a_ne^{-x_n}$

That can't be right though....

2. Hello,

Hey... I think you have serious problems with the definitions... f is the pdf, the random variable does not equal $e^{-x}$

3. So what would Y equal to? Would I need to do a change of variable?

4. Originally Posted by statmajor
Let Y1<Y2<...<Yn be the order statistics of a random sample of size n from the pdf $f(x) = e^{-x}$ x ranging from 0 to infinity.

Demonstrate that all linear functions of Y1, Y2,...,Yn such as $\Sigma a_i Y_i$ can be expressed as a linear function of independent random variables.

so:

$\Sigma a_i Y_i = a_1Y_1 + a_2Y_2 + ... + a_nY_n = a_1e^{-x_1} + a_2e^{-x_2} +...+a_ne^{-x_n}$

That can't be right though....
$Y_1, Y_2, \dots , Y_n$ are the order statistics of an Exponential(1) random variable; so $W_1 = Y_1, \; W_2 = Y_2 - Y_1, \; W_3 = Y_3 - Y_2, \dots , W_n = Y_n - Y_{n-1}$ are n independent (exponentially distributed) random variables.
A proof of this fact can be found in Feller, "An Introduction to Probability Theory and Its Applications, Volume II"; or you may be familiar with the statement that if arrival times are exponentially distributed then the inter-arrival times are independent and exponentially distributed.

Then
$Y_1 = W_1$
$Y_2 = W_2 + W_1$
$Y_3 = W_3 + W_2 + W_1$
etc.,
so the Y's are linear functions of the W's. Hence any linear function of the Y's can be re-written as a linear function of the W's.

5. God, I'm such an idiot. There was another part to this question where it asks me to prove that Z1 = nY1 Z2 = (n-1)(Y2 - Y1) Z3 = (n-2)(Y3-Y2),...,Zn = Yn - Y(n-1)

so:

$\Sigma a_i Y_i = a_1\frac{Z_1}{n} + a_2(\frac{Z_2}{n-1} + \frac{Z_1}{n})+..+a_n(Z_n + ... + \frac{Z_1}{n})$

Can't believe I didn't realise this sooner. Thanks a lot.

Last question: is the nth term correct?

6. The last equation looks consistent with your definition of the Zs.

7. Thanks for all your help.