1. ## random variable convergence, need help please

Wn has density [1+(x/n)]/[1+(0.5n)] for 0<x<1, and 0 otherwise. The random variable W has a uniform distribution on [0,1]. Prove that {Wn} converges in distribution to W.

I have so far that
mWn(s)=E(e^(sWn))= integral (0 to 1): e^(sWn)*((1+(x/n)))/(1+(1/2n))dWn
= ((1+(x/n)))/(1+(1/2n)) * integral (0 to 1): e^(sWn)dWn
which equals (as lim n-> inf.)
integral (0 to 1): e^(Sw)dw

mW(s)=E(e^(Sw))=integral (0 to 1): e^(Sw)dw

But now how do I show the convergence?

Any help is appriciated.

2. IT's hard to follow.
It would be nice if you typed in tex, that's why I couldn't understand your last post

BUT [1+(x/n)]/[1+(0.5n)] isn't a density, one of the n's are wrong.

And forget MGFs, just get the cummulative distribution and show they converge.

3. i doubled checked the question in my book, and its exactly what i typed.... maybe its some different kind of density function?

Anyways i got the c.d.f. of both random variables to match, but is that all i do for this?

4. Originally Posted by Sneaky
i doubled checked the question in my book, and its exactly what i typed.... maybe its some different kind of density function?

Anyways i got the c.d.f. of both random variables to match, but is that all i do for this?
If I were a betting man I would put money on it actually saying the equivalent of:

[1+(x/n)]/[1+1/(2n)] for 0<x<1, and 0 otherwise

CB

5. what?

6. one of the n's are wrong.

$\displaystyle \int_0^1[1+(x/n)]dx=[1+{1\over 2n}]$

7. Originally Posted by Sneaky
what?
That makes it a density and consistent with what you actually use in your third line.

CB

8. Originally Posted by matheagle
one of the n's are wrong.

$\displaystyle \int_0^1[1+(x/n)]dx=[1+{1\over 2n}]$
so if i am understanding this right, for the cdf of Wn, as n approaches infinity, the cdf will become 1. But for the cdf of W, its just

0 if w<=0
w 0<w<1
1 w>= 1

I don't fully understand why Wn converges in distribution to W.

9. Originally Posted by Sneaky
Wn has density [1+(x/n)]/[1+(0.5n)] for 0<x<1, and 0 otherwise. The random variable W has a uniform distribution on [0,1]. Prove that {Wn} converges in distribution to W.

I have so far that
mWn(s)=E(e^(sWn))= integral (0 to 1): e^(sWn)*((1+(x/n)))/(1+(1/2n))dWn
= ((1+(x/n)))/(1+(1/2n)) * integral (0 to 1): e^(sWn)dWn
which equals (as lim n-> inf.)
integral (0 to 1): e^(Sw)dw

mW(s)=E(e^(Sw))=integral (0 to 1): e^(Sw)dw

But now how do I show the convergence?

Any help is appriciated.
For $\displaystyle x\le 1$ put

$\displaystyle \displaystyle F_n(x)=\int_{\xi=0}^x \dfrac{1+\frac{\xi}{n}}{1+\frac{1}{2n}}\ d\xi$

$\displaystyle \displaystyle F_(x)=\int_{\xi=0}^x 1 \ d\xi$

You need to show that:

$\displaystyle F_n(x) \to F(x)$

CB

10. just integrate, calc 1, and show that the limit of Fn approaches x for 0<x<1

11. after integration i get

x^2 + 2nx
----------- -> x
2n +1

I'm not sure what to do to the left hand side, since subbing n as infinity here will not work.

EDIT: i factored out x, then the 2n+1 cancelled out, and was just left with x, so x -> x, so yeah i think i got it

12. this is basic calc 1

$\displaystyle {x+{x^2\over 2n}\over 1+{1\over 2n}} =x\left({1+{x\over 2n}\over 1+{1\over 2n}}\right)\to x$

13. i'm not sure where you got those equations

14. Originally Posted by Sneaky
i'm not sure where you got those equations
1. Quote what you are referring to, saves us guessing.

2. Presumably you have done calculus and so are familiar with convergence of sequences?

CB