Probability - transformation of a random variable

[dizzle1518]

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.

 

[dizzle1518]

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.

 

[dizzle1518]

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.You didn't say what you asked your teacher, so I don't what her answer meant.

 

[dizzle1518]

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?

You didn't say what you asked your teacher, so I don't what her answer meant.

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 that's 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 that's 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.

 

[dizzle1518]

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.

 

[dizzle1518]

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]

Y =1 exactly when X is between 0 and 1. What's the probability that the normal random variable X with mean 0 and standard deviation 1 is between 0 and 1? (It's about .3413.) The answer to that gives you the probability that Y = 1.