I just solved this. Can you? (MX2 students should have a go) (1 Viewer)

nightweaver066 · Apr 12, 2012

Shadowdude said:
Wow, forgot about this thread.

Yes, that's on the right track definitely.

How? Could you show working? I think you might've taken a limit wrong or something.

Limits of...?

Working is a bit long, but i combined the x and x^(alpha - 1), took out the constants (the T(alpha) and beta thing), considered the integral seperately, applied tabular integration, deduced what the integral would result in:
$\alpha!(-1)^{\alpha + 2}\beta^{\alpha + 1}$

Subbed that back in to the integral, simplified and got $\alpha \beta (-1)^{\alpha}$

Edit: Nevermind, figured it out. I forgot to include something while integrating and now i have the answer.

bleakarcher · Apr 12, 2012

I got:

lim[R->infinity] I(A)=lim[R->infinity] AB*I(A-1) where A is alpha and B is beta.

?

bleakarcher · Apr 12, 2012

I give up with this problem...

jet · Apr 12, 2012

Are there any restrictions on $\alpha$ and $\beta$

Shadowdude · Apr 12, 2012

bleakarcher said:
I got:

lim[R->infinity] I(A)=lim[R->infinity] AB*I(A-1) where A is alpha and B is beta.

?

You don't take limits, you keep working it out. I(A) and I(A-1) are difficult to work with, so you simplify it down.

jetblack2007 said:
Are there any restrictions on $\alpha$ and $\beta$

As in...? I don't think so... or at least when I did the problem, I didn't think of that.

jet · Apr 12, 2012

Shadowdude said:
You don't take limits, you keep working it out. I(A) and I(A-1) are difficult to work with, so you simplify it down.

As in...? I don't think so... or at least when I did the problem, I didn't think of that.

Care to post a solution?

nightweaver066 · Apr 12, 2012

jetblack2007 said:
Care to post a solution?

$E(X)=\lim_{R \to \infty}\int_{0}^{R} x\frac{e^{-x/\beta}x^{\alpha - 1}}{\Gamma (\alpha) \beta^{\alpha}} dx$

$= \frac{1}{(\alpha - 1)!\beta^{\alpha}} \times \lim_{R \to \infty}e^{-\frac{x}{\beta}}x^{\alpha}dx$

$After applying tabular integration to the integral with the limit,$

$\lim_{R \to \infty} \int^R_0 e^{-\frac{x}{\beta}}x^{\alpha}dx$

$= [x^{\alpha} \times (-\beta e^{-x/\beta} - \alphax^{\alpha + 1} \times (\beta^2e^{-x/\beta}) - ... + \alpha! \times (-1)^{\alpha}(-1)^{\alpha + 1}\beta^{\alpha + 1}e^{-x/\beta}]^{\infty}_0$

$$As x $ \rightarrow \infty $ , e^{-x/\beta} \rightarrow 0, \therefore \phi e^{-x/\beta} \rightarrow 0$

$where \phi is a real number$

$\therefore \lim_{R \to \infty} \int^R_0 e^{-\frac{x}{\beta}}x^{\alpha}dx = -\alpha! (-1)^{2\alpha + 1}\beta^{\alpha + 1}$

$= -\alpha! (-1)^{2\alpha + 1} \beta^{\alpha + 1}$

$= \alpha! (-1)^{2\alpha + 2} \beta^{\alpha + 1}$

$= \alpha! \beta^{\alpha + 1}$

$\therefore E(x) = \frac{1}{(\alpha - 1)!\beta^{\alpha}} \times \alpha! \beta^{\alpha + 1}$

$= \alpha \beta$

RealiseNothing · Apr 12, 2012

bleakarcher said:
I got:

lim[R->infinity] I(A)=lim[R->infinity] AB*I(A-1) where A is alpha and B is beta.

?

That's what I'm up to.

@shadowdude or some one who did the question, have I nearly got the answer or is there still a long way to go?

jet · Apr 12, 2012

That only works if $\alpha$ is a natural number. So unless shadowdude has left something out, that isn't the solution.

nightweaver066 · Apr 12, 2012

jetblack2007 said:
That only works if $\alpha$ is a natural number. So unless shadowdude has left something out, that isn't the solution.

Because of the $\alpha!$ ?

jet · Apr 12, 2012

Because the gamma function equals (alpha - 1)! only when alpha is a natural number non-zero. Otherwise it's defined by a complex integral expression.

Shadowdude · Apr 12, 2012

RealiseNothing said:
That's what I'm up to.

@shadowdude or some one who did the question, have I nearly got the answer or is there still a long way to go?

That's close. Bit more to go.

jetblack2007 said:
That only works if $\alpha$ is a natural number. So unless shadowdude has left something out, that isn't the solution.

Ah, yeah okay... now I know what you mean. Alpha and beta are natural numbers. I think I implied it but didn't specify it - and the question from my book doesn't specify it either, but... I'll say they are natural numbers, so it'll work.

Anyways, here's how I did it:

$E(X)=\lim_{R \to \infty}\int_{0}^{R} x\frac{e^{-x/\beta}x^{\alpha - 1}}{\Gamma (\alpha) \beta^{\alpha}} dx$

Factor out the constants and combine the powers of x:

$E(X)=\frac{1}{{\Gamma (\alpha) \beta^{\alpha}}}\lim_{R \to \infty}\int_{0}^{R} e^{-x/\beta}x^{\alpha} dx$

Let:

$I_{\alpha} = \lim_{R \to \infty} \int_{0}^{R} e^{-x/\beta}x^{\alpha} dx$

So we integrate by parts:

$\int u v' = uv - \int vu'$

where:

$u = x^{\alpha} \Rightarrow u' = \alpha x^{\alpha - 1}$
$v' = e^{-x/ \beta} \Rightarrow v = -\beta e^{-x/ \beta}$

Then:

$I_{\alpha} = \lim_{R \to \infty} \left [ -\beta x^{\alpha}e^{-x/\beta} \right ]_{0}^{R} - \lim_{R \to \infty} \int_{0}^{R} \left ( -\beta e^{-x/ \beta} \right ) \left (\alpha x^{\alpha - 1} \right ) \;dx$

$I_{\alpha} = \lim_{R \to \infty} \left (-\beta R^{\alpha}e^{-R/\beta} \right ) + \lim_{R \to \infty} \alpha \beta \int_{0}^{R} e^{-x/ \beta} \right )x^{\alpha -1} \;dx\\$

Now, using the second given result:

$\lim_{R \to \infty} \left (-\beta R^{\alpha}e^{-R/ \beta} \right ) = -\beta \lim_{R \to \infty} \left (R^{\alpha}e^{-R/ \beta} \right ) = -\beta (0) = 0$

So our expression simplifies to:

$I_{\alpha} = \lim_{R \to \infty} \alpha \beta \int_{0}^{R} e^{-x/ \beta} \right )x^{\alpha -1} \;dx = \alpha \beta I_{\alpha - 1}$

So as we have:

$I_{\alpha} = \alpha \beta \;I_{\alpha - 1}$

We can see that:

$I_{\alpha} = (\alpha (\alpha - 1)(\alpha - 2)... 2) \beta^{\alpha - 1}\; I_{1}$

We solve this manually:

$I_{\alpha} = (\alpha !) \beta^{\alpha - 1} \lim_{R \to \infty} \left (\int_{0}^{R} x\, e^{-x/\beta} \;dx \right )$

By integration by parts, as seen above:

$I_{\alpha} = (\alpha !) \beta^{\alpha - 1} \lim_{R \to \infty} \left ( \left [ -x\beta e^{-x/\beta} \right ]_{0}^{R} + \beta \int_{0}^{R}e^{-x/\beta} \; dx \right ) \right ) \right )$

$I_{\alpha} = (\alpha !) \beta^{\alpha - 1} \lim_{R \to \infty} \left (-R\beta e^{-R/\beta} + \beta \left [-\beta e^{-x/\beta} \right ]_{0}^{R} \right ) \right )$

$I_{\alpha} = (\alpha !) \beta^{\alpha - 1} \lim_{R \to \infty} \left (-R\beta e^{-R/\beta} - \beta ^{2} \left (e^{-R/\beta} - e^{-0/\beta} \right ) \right ) \right )\\$

$I_{\alpha} = (\alpha !) \beta^{\alpha - 1} \lim_{R \to \infty} \left (-R\beta e^{-R/\beta} - \beta ^{2} \left (e^{-R/\beta} - 1 \right ) \right ) \right )\\$

$I_{\alpha} = (\alpha !) \beta^{\alpha - 1} (0 - \beta^{2}(-1)) = \alpha!\, \beta^{\alpha + 1}$

Recall that:

$E(X)=\frac{1}{{\Gamma (\alpha) \beta^{\alpha}}} I_{\alpha}$

So:

$E(X)=\frac{1}{{\Gamma (\alpha) \beta^{\alpha}}} \alpha ! \beta^{\alpha + 1}$

Use given fact one:

$E(X)=\frac{\alpha ! \; \beta^{\alpha + 1}}{{(\alpha - 1)! \beta^{\alpha}}} = \alpha \beta$

i hope that's right. my head hurts and i'd hate to see that i fudged the answer. <_<

Shadowdude · Apr 12, 2012

So I just realised that the question didn't specify alpha and beta to be natural numbers because... they don't have to be.

So umm, I have to redo this question.

FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF

Carrotsticks · Apr 12, 2012

Shadowdude said:
So I just realised that the question didn't specify alpha and beta to be natural numbers because... they don't have to be.

So umm, I have to redo this question.

FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF

Alpha was defined in the Gamma function, which already implies it being a Natural number (I doubt this question assumed knowledge of non-integer factorials..)

Beta was not explicitly defined to be an integer but we did nothing with it that required it to be an integer, so it's okay.

bleakarcher · Apr 13, 2012

ah shit, thats what i did wrong. I let I(A) represent just the integral without the limit...

Shadowdude · Apr 13, 2012

Carrotsticks said:
Alpha was defined in the Gamma function, which already implies it being a Natural number (I doubt this question assumed knowledge of non-integer factorials..)

Beta was not explicitly defined to be an integer but we did nothing with it that required it to be an integer, so it's okay.

I dunno, the... stats course I'm in isn't taught super well.

When we learned the gamma distribution it was like "Oh btw, there's this thing called the gamma function - here's what it is, here's three properties... and now this is the gamma distribution, that's the formula. Now the next distribution we have is..."

And then you know how tutorial problems have absolutely nothing to do with what's taught in lectures.

Maybe tomorrow I'll re-think what's going on and see if I actually do need to re-do the question, but you do raise a point.

bleakarcher said:
ah shit, thats what i did wrong. I let I(A) represent just the integral without the limit...

you can do it that way, but you'll just have a whole lot of unnecessary terms when you sub it back in - all of which will be zero

jet · Apr 13, 2012

Carrotsticks said:
Alpha was defined in the Gamma function, which already implies it being a Natural number (I doubt this question assumed knowledge of non-integer factorials..)

Beta was not explicitly defined to be an integer but we did nothing with it that required it to be an integer, so it's okay.

Actually, the question just states that the gamma function is defined as (alpha - 1)! when alpha is natural and non-zero, there's nothing there that specifically restricts alpha to be natural.

Shadowdude · Apr 13, 2012

jetblack2007 said:
Actually, the question just states that the gamma function is defined as (alpha - 1)! when alpha is natural and non-zero, there's nothing there that specifically restricts alpha to be natural.

The original original original question is:

Let X ~ Gamma(alpha, beta). Prove E(X) = (alpha)(beta)

And then in that, you get an integral which I simplified and presented here... and I'm now 85% sure I did something wrong. dammit.

jet · Apr 13, 2012

Well if you learn about the Gamma distribution with alpha natural and non-zero, then you're fine. Otherwise, back to the drawing board.

bleakarcher · Apr 13, 2012

As I was falling asleep yesterday I realised what I did wrong. In setting up the reduction formula, I didn't take out the gamma function and the beta to the alpha LOL...shit.

I just solved this. Can you? (MX2 students should have a go) (1 Viewer)

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