# Probability - transformation of a random variable

In an analog to digital conversion and analog waveform is sampled, quantized and coded. A quantized function is a function that assigns to each sample value x a value y from a generally finite set of predetermined values. Consider the quantized defined by g(x)=[x]+1, where [x] denotes the greatest integer less than or equal to x. Suppose that x has a standard normal distribution and pit Y=g(x). Specify the distribution of Y. Ignore values of Y for which the probability is essentially zero.

Going by how the book taught it I would start this problem by computing the inverse of g(x). However this function has no inverse. Any suggestions how to proceed?

Stephen Tashi
What's the probability that Y = 1? For that to happen, X must be in [0,1). You can compute the probability of X being in that interval by using the normal distribution.

thanks. looks to me like the interval would be (-infinity, y-1). so the distribution of why would be $$\Phi(y-1)$$. is this right?

Stephen Tashi
thanks. looks to me like the interval would be (-infinity, y-1). so the distribution of why would be $$\Phi(y-1)$$. is this right?

Are you taking about the interval to use when computing the cumulative distribution of Y? Yes, that's right if you're using $$\Phi$$ to denote the cumulative normal.

The wording of the problem indicates that you should specify a finite list of integers on which the density of Y is non-zero.

i think it just asks for the distribution of Y. How would you know which integers to specify anyway?

I e-mailed my teacher and she replied with the below:

"No, for every y the corresponding interval for x has to be (-infinity,[y]). The endpoint should be an integer."

Does she mean it should be Fy=phi(y)??

Stephen Tashi
How would you know which integers to specify anyway?

There can't be that many integers with a significant probability. The standard normal has standard deviation = 1. There isn't much chance of getting Y = 12.

There can't be that many integers with a significant probability. The standard normal has standard deviation = 1. There isn't much chance of getting Y = 12.

got you. then picking up to what numbers you want to define the distribution of Y is up to the person solving the problem?

i asked her the same thing, i.e. if the interval was (-infinity, y-1) and if the distribution of Y is phi(y-1), and thats what she answered

Stephen Tashi
got you. then picking up to what numbers you want to define the distribution of Y is up to the person solving the problem?

It's up to them to do the work. There won't be much disagreement on which integers have non-zero probability if they all use similar tables of the normal distribution.

i asked her the same thing, i.e. if the interval was (-infinity, y-1) and if the distribution of Y is phi(y-1), and thats what she answered

The phrase "if the interval was (-infinity, y-1)" isn't a complete sentence. So I assume she ignored it.

The answer to "if the distribution of Y is phi(y-1)" is no. She answered correctly. The distribution of Y isn't defined on all real numbers, only on integers. Her answer indicates that you only use the calculation on integer values of Y.

so what would be the distribution of Y then?

Stephen Tashi
In my opinion, you are trying to give an abstract answer to a problem that wants you to do some numerical work. It wants a list of integers and their probability of occurrence. Are you in doubt about how to compute the numerical probability that Y = 1 or Y = -1 ? I think the problem wants you give the probability density for Y, not the cumulative distribution.

You have the correct idea about using $$\phi$$. It's the imprecision in your statement of that idea that your teacher objects to. However, I don't think stating the formula for the cumulative distribution of Y is what the problem requests.

Are you in doubt about how to compute the numerical probability that Y = 1 or Y = -1 ?

a bit yes

I think the problem wants you give the probability density for Y, not the cumulative distribution.

would Y have a density or mass function? i thought mass since Y is defined over integers only. the fact that X is continuous and Y discrete is throwing me off somewhat.

Stephen Tashi