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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: sxcalexc on August 30, 2008, 07:29:43 pm

Title: Continuous Random Variable Question
Post by: sxcalexc on August 30, 2008, 07:29:43 pm: Hey guys this is the last question in the exercise... I'm having a bit of difficulty.. Any help appreciated, thank you.
(http://img253.imageshack.us/img253/8324/47267503hn0.jpg)
Title: Re: Continuous Random Variable Question
Post by: Collin Li on August 30, 2008, 07:49:36 pm: a)

$\frac{d}{dx}\left(kxe^{-kx}\right) = ke^{-kx} - k^2xe^{-kx}$

Anti-differentiate both sides now, to yield:

$\implies kxe^{-kx} = \int \left(ke^{-kx} - k^2xe^{-kx}\right)\,dx$

Cancelling out a $k$ : $\implies xe^{-kx} = \int \left(e^{-kx} - kxe^{-kx}\right)\, dx$

Splitting the integral term by term: $\implies xe^{-kx} = \int e^{-kx}\,dx - \int kxe^{-kx}\, dx$

Rearranging to obtain the desired integral: $\implies \int kxe^{-kx}\, dx = \int e^{-kx}\,dx - xe^{-kx}$

$\implies \int kxe^{-kx}\, dx = -\frac{1}{k}e^{-kx} - xe^{-kx} = -e^{-kx}\left(\frac{1}{k} + x\right)$

b)

Mean:

$\mbox{E}[X] = \int_0^\infty x\cdot\frac{1}{\lambda}e^{-\frac{x}{\lambda}}\,dx$

Let $k = \frac{1}{\lambda}$

Using part (a):

$\implies \mbox{E}[X] = \int_0^\infty kxe^{-kx}\,dx = \left[-e^{-kx}\left(\frac{1}{k} + x\right)\right]^\infty_0$

Evaluating the integral: lots of terms go to zero, except 2. One is complicated and one is a constant:

$\implies \mbox{E}[X] = \lim_{x\to\infty}\left(-xe^{-kx}\right) + \frac{1}{k}$

Fact: $\lim_{x\to\infty}\left(-xe^{-kx}\right) = 0$

This fact holds because the exponential function decreases far faster than the linear function, hence as $x$ goes to infinity, this becomes zero. That's not a solid proof, but it is a general rule of thumb that works -- exponentials change faster than polynomials. This isn't something you need to know for Methods - so it's the exercise's problem for having that ambigious term there.

Since $k = \frac{1}{\lambda}$

$\implies \mbox{E}[X] = \frac{1}{k} = \lambda$

c)

Graph-sketching problem involving exponential functions. Shouldn't be too difficult. Just make sure you show the effects of a different size of $\lambda$ .

d)

Larger values of $\lambda$ will result in smaller $\frac{1}{\lambda}$ , which decrease the vertical dilation of the graph, so the curves with larger values of $\lambda$ will be flatter. The horizontal dilation of the graph will increase with larger values of $\lambda$ , which also contributes to a flatter-looking graph.

This is consistent with the definition found in part (b), $\mbox{E}[X] = \lambda$ . Since the graph starts at $x=0$ , and it diminishes rapidly (being a decaying exponential function), a flatter graph that takes longer to decay (less steep) is required to have a larger mean. Hence, a larger mean (larger value of $\lambda$ ) is consistent with a flatter graph.
Title: Re: Continuous Random Variable Question
Post by: danieltennis on August 30, 2008, 07:51:34 pm: Coblin, what program did u use to write all these formulas.
Title: Re: Continuous Random Variable Question
Post by: Collin Li on August 30, 2008, 07:54:42 pm: Quote from: danieltennis on August 30, 2008, 07:51:34 pm
Coblin, what program did u use to write all these formulas.

LaTeX, something that is embedded into this forum. I just have to type in a particular format in between TEX tags, and it will come out in a nicer looking viewer format (as a picture). To learn how to use it, see this thread: http://vcenotes.com/forum/index.php/topic,3137.0.html
Title: Re: Continuous Random Variable Question
Post by: danieltennis on August 30, 2008, 07:57:09 pm: Quote from: coblin on August 30, 2008, 07:54:42 pm
Quote from: danieltennis on August 30, 2008, 07:51:34 pm
Coblin, what program did u use to write all these formulas.

LaTeX, something that is embedded into this forum. I just have to type in a particular format in between TEX tags, and it will come out in a nicer looking viewer format (as a picture). To learn how to use it, see this thread: http://vcenotes.com/forum/index.php/topic,3137.0.html
thanks
Title: Re: Continuous Random Variable Question
Post by: sxcalexc on August 30, 2008, 10:17:25 pm: Thx very much for your effort Coblin. In regards to part a, the book have this as the answer for the derivative and integral, respectively:
(http://img84.imageshack.us/img84/5606/hmmdv1.jpg)
Do you think they just left out the second half of the derivative? The integral is the same.
EDIT: Actually the integral is slightly different. One of the signs differs. I think they got that wrong too ? ???
Title: Re: Continuous Random Variable Question
Post by: Collin Li on August 30, 2008, 10:41:58 pm: Yeah, I think they got that wrong, haha. It's not a well designed question (and answer), coz it had to use that fact that isn't in the course, in part (b).
Title: Re: Continuous Random Variable Question
Post by: sxcalexc on August 30, 2008, 10:48:28 pm: Quote from: coblin on August 30, 2008, 10:41:58 pm
Yeah, I think they got that wrong, haha. It's not a well designed question (and answer), coz it had to use that fact that isn't in the course, in part (b).
Right.. what bastards. Lol, I kind of figured that 'rule of thumb' applied just by looking at it logically anyway. Perhaps that's what they want us to do? Nevertheless, muchas gracias.
Title: Re: Continuous Random Variable Question
Post by: Collin Li on August 30, 2008, 10:53:02 pm: Yeah, Ahmad did say to me, looking at it like: $xe^{-x} = \frac{x}{e^x}$ is obvious, but it still looks like $\frac{\infty}{\infty}$ (indeterminate) when you let $x \to \infty$
Title: Re: Continuous Random Variable Question
Post by: sxcalexc on August 30, 2008, 10:54:50 pm: Quote from: coblin on August 30, 2008, 10:53:02 pm
Yeah, Ahmad did say to me, looking at it like: $xe^{-x} = \frac{x}{e^x}$ is obvious, but it still looks like $\frac{\infty}{\infty}$ (indeterminate) when you let $x \to \infty$

Mmm.. indeed. Duly noted. So, basically, whenever something is put to a power of a negative infinity, so to speak, it is effectively 0.

Ah, another question for you. In part a where you cancelled the k. Can you do that in the integrals as well? I was told that you can't multiply or divide like that inside integrals.
Title: Re: Continuous Random Variable Question
Post by: Glockmeister on August 30, 2008, 11:24:35 pm: I think I remember doing this question actually, had to ask my teacher about it.
Title: Re: Continuous Random Variable Question
Post by: Mao on August 30, 2008, 11:28:23 pm: if you must, L'Hopital's rule (taking derivative of top and bottom when encountering an indeterminant)

$\lim_{x\to \infty} \frac{x}{e^x} = \lim_{x\to \infty} \frac{1}{e^x} = \frac{1}{\infty} = 0$

however, you don't learn this until first year uni :P
Title: Re: Continuous Random Variable Question
Post by: Collin Li on August 30, 2008, 11:39:48 pm: Quote from: sxcalexc on August 30, 2008, 10:54:50 pm
Ah, another question for you. In part a where you cancelled the k. Can you do that in the integrals as well? I was told that you can't multiply or divide like that inside integrals.

Since $k$ is a constant, I can just take it out of the integral, and it will cancel out.
Title: Re: Continuous Random Variable Question
Post by: Matt The Rat on August 31, 2008, 12:29:17 am: Quote from: Mao on August 30, 2008, 11:28:23 pm

$\lim_{x\to \infty} \frac{x}{e^x} = \lim_{x\to \infty} \frac{1}{e^x} = \frac{1}{\infty} = 0$

Writing $\frac{1}{\infty} = 0$ lost my mate marks on a MUEP assignment.
Title: Re: Continuous Random Variable Question
Post by: Captain on September 03, 2008, 08:48:51 pm: Quote from: Matt The Rat on August 31, 2008, 12:29:17 am
Quote from: Mao on August 30, 2008, 11:28:23 pm

$\lim_{x\to \infty} \frac{x}{e^x} = \lim_{x\to \infty} \frac{1}{e^x} = \frac{1}{\infty} = 0$

Writing $\frac{1}{\infty} = 0$ lost my mate marks on a MUEP assignment.

That's because ${\infty}$ isn't a number.
But,
$\lim_{x\to \infty} \frac{x}{e^x} = \lim_{x\to \infty} \frac{1}{e^x} = 0$

So I agree with you. Mao is wrong :) Haha

[And no Mao, I will not prove L'Hopital's rule]