Uniform Distribution over n and its limit

[JonoPUH]

Homework Statement

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

Homework Equations

So Y_n has c.d.f Y_n(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 Y_n 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 Y_n/n clearly tends to 0 as n → 0 since we have 1/n.

Thanks

 

[Ray Vickson]

Homework Statement

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

Homework Equations

So Y_n has c.d.f Y_n(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 Y_n 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 Y_n/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 $$F_n(x) = P(Y_n \leq x) = \left\{<br /> \begin{array}{cl}\frac{\lfloor x \rfloor}{n},& \: 0 \leq x < n\\<br /> 1,&\: x \geq n<br /> \end{array}\right.$$ as you said. The distribution of Y_n/n is
$$G_n(x) = P\left( \frac{Y_n}{n} \leq x \right) = P( Y_n \leq n x ) = F_n(nx)<br /> = \left\{ \begin{array}{cl}<br /> \frac{\lfloor nx \rfloor}{n}, &\: 0 \leq nx < n\\<br /> 1, & \: nx \geq n<br /> \end{array} \right.$$

 

[JonoPUH]

Thank you for your clarification on notation and help!