P.M.F. with Decimal Expansion

Jan 2014
103
3
Arizona
Define the random variable Y on [0, 1] as follows. Pick a number y at random, and take its nth root.
Let Y be the first digit in the decimal expansion of y1/n.
E.g., Y = 3 when y1/n is in [.3, .4), etc.
Find the probability mass function P(Y = k); k = 0,...,9.

First of all, I'm confused since Y is on [0, 1]. Doesn't that mean that Y can only take on values from 0 to 1?
So why is it in the example that Y = 3?
Also, when we pick a number y, does that mean any number y in R?

And then from there, I'm not quite sure how to approach this problem.
 

romsek

MHF Helper
Nov 2013
6,725
3,030
California
Define the random variable Y on [0, 1] as follows. Pick a number y at random, and take its nth root.
Let Y be the first digit in the decimal expansion of y1/n.
E.g., Y = 3 when y1/n is in [.3, .4), etc.
Find the probability mass function P(Y = k); k = 0,...,9.

First of all, I'm confused since Y is on [0, 1]. Doesn't that mean that Y can only take on values from 0 to 1?
So why is it in the example that Y = 3?
Also, when we pick a number y, does that mean any number y in R?

And then from there, I'm not quite sure how to approach this problem.
This is terribly worded.

Y is not on [0,1] as far as I can tell. Y is {0,1,2,3,4,5,6,7,8,9}.

What I think they mean is that the original $y \in [0,1]$

So $y$ is a uniform rv on $[0,1]$

$Y$ is the first digit in the decimal expansion of $y^{1/n}$

Before you do any calculations what does your intuition tell you about what the PMF of $Y$ should be?
 
Jan 2014
103
3
Arizona
Well I'm not too familiar with decimal expansions and I still don't know if I understand what's going on exactly,
but I would say perhaps P(Y = k) = 1/10 for k = 0, 1, .., 9 since it seems that each value should be equally likely.
 

romsek

MHF Helper
Nov 2013
6,725
3,030
California
Well I'm not too familiar with decimal expansions and I still don't know if I understand what's going on exactly,
but I would say perhaps P(Y = k) = 1/10 for k = 0, 1, .., 9 since it seems that each value should be equally likely.
This is what's going on.

$y$ is a random variable that is Uniform [0,1].

$\hat{y} = y^{1/n}$

note that $\hat{y}\in [0,1]$

$Y=\lfloor 10 \hat{y} \rfloor$

I originally thought that the PMF of Y would be uniform but I discovered it depends on $n$

We need to find the CDF of $y^{1/n}$

can you do that?

Once we have that we can evaluate the probability of each of the regions corresponding to the values of $Y$.