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

**jet** · 12 Apr 2012, 11:52 PM

Originally Posted by Shadowdude

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** · 13 Apr 2012, 12:05 AM

Originally Posted by jetblack2007

Care to post a solution?

Originally Posted by jetblack2007

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** · 13 Apr 2012, 12:07 AM

Originally Posted by bleakarcher

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** · 13 Apr 2012, 12:12 AM

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

**nightweaver066** · 13 Apr 2012, 12:19 AM

Originally Posted by jetblack2007

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** · 13 Apr 2012, 12:21 AM

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** · 13 Apr 2012, 12:41 AM

Originally Posted by RealiseNothing

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.

Originally Posted by jetblack2007

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** · 13 Apr 2012, 12:53 AM

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** · 13 Apr 2012, 12:56 AM

Originally Posted by Shadowdude

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** · 13 Apr 2012, 1:01 AM

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

**Shadowdude** · 13 Apr 2012, 1:02 AM

Originally Posted by Carrotsticks

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.

Originally Posted by bleakarcher

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** · 13 Apr 2012, 1:07 AM

Originally Posted by Carrotsticks

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** · 13 Apr 2012, 1:09 AM

Originally Posted by jetblack2007

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** · 13 Apr 2012, 1:10 AM

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** · 13 Apr 2012, 1:24 PM

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.

**RealiseNothing** · 13 Apr 2012, 3:58 PM

Originally Posted by bleakarcher

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.

How did you possibly use integration by parts with them still in the integral lol.

**bleakarcher** · 13 Apr 2012, 4:05 PM

LOL, they're just constants...

**RealiseNothing** · 13 Apr 2012, 4:08 PM

Originally Posted by bleakarcher

LOL, they're just constants...

I know, but it would of been very annoying doing it with them still in the integral lol.

**bleakarcher** · 13 Apr 2012, 4:11 PM

yeh, i guess. I should have taken them out like a normal person lol. Because they were part of the reduction formula they stayed in the reduction formula and so they never cancelled the alpha factorial and the beta^(alpha+1), which was the answer I ended up with.

**RealiseNothing** · 13 Apr 2012, 4:11 PM

Originally Posted by Shadowdude

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

Yer that's the exact same way I did it. It wasn't that difficult of a question imo, just looks very daunting but once you take out the constants and apply reduction method it becomes pretty simple.

Would a question like this appear in a MX2 exam?

**bleakarcher** · 13 Apr 2012, 4:13 PM

^ probably, it'd be in parts though.

**RealiseNothing** · 13 Apr 2012, 4:14 PM

Originally Posted by bleakarcher

^ probably, it'd be in parts though.

haha good pun (not sure if it was intended though).

**bleakarcher** · 13 Apr 2012, 4:15 PM

LOL, naa it wasnt.

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