Uniform Distribution over n and its limit

JonoPUH
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Homework Statement



Let Yn be uniform on {1, 2, . . . , n} (i.e. taking each value with probability 1/n). Draw the distribution function of Yn/n. Show that the sequence Yn/n converges in distribution as n → ∞. What is the limit?

Homework Equations



So Yn has c.d.f Yn(x) = |x|/n where |x| is the nearest integer less than x.

The Attempt at a Solution



Is it ok for me to just divide the c.d.f of Yn by n, and so plot |x|/n2?
This seems far too easy for 2nd year of university (I really do not trust myself with probability) and also it only gives a total for x > n of 1/n. Is this allowed?

If so, then Yn/n clearly tends to 0 as n → 0 since we have 1/n.

Thanks
 
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JonoPUH said:

Homework Statement



Let Yn be uniform on {1, 2, . . . , n} (i.e. taking each value with probability 1/n). Draw the distribution function of Yn/n. Show that the sequence Yn/n converges in distribution as n → ∞. What is the limit?

Homework Equations



So Yn has c.d.f Yn(x) = |x|/n where |x| is the nearest integer less than x.

The Attempt at a Solution



Is it ok for me to just divide the c.d.f of Yn by n, and so plot |x|/n2?
This seems far too easy for 2nd year of university (I really do not trust myself with probability) and also it only gives a total for x > n of 1/n. Is this allowed?

If so, then Yn/n clearly tends to 0 as n → 0 since we have 1/n.

Thanks

Your notation is horrible and will get you into trouble. Yn is a random variable, not a function of x. Its distribution function is F_n(x) = P(Y_n \leq x) = \left\{<br /> \begin{array}{cl}\frac{\lfloor x \rfloor}{n},&amp; \: 0 \leq x &lt; n\\<br /> 1,&amp;\: x \geq n<br /> \end{array}\right. as you said. The distribution of Y_n/n is
G_n(x) = P\left( \frac{Y_n}{n} \leq x \right) = P( Y_n \leq n x ) = F_n(nx)<br /> = \left\{ \begin{array}{cl}<br /> \frac{\lfloor nx \rfloor}{n}, &amp;\: 0 \leq nx &lt; n\\<br /> 1, &amp; \: nx \geq n<br /> \end{array} \right.
 
Thank you for your clarification on notation and help!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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