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Probability - transformation of a random variable

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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?
 

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  • #2
Stephen Tashi
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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.
 
  • #3
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thanks. looks to me like the interval would be (-infinity, y-1). so the distribution of why would be [tex]\Phi(y-1)[/tex]. is this right?
 
  • #4
Stephen Tashi
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thanks. looks to me like the interval would be (-infinity, y-1). so the distribution of why would be [tex]\Phi(y-1)[/tex]. 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 [tex]\Phi [/tex] 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.
 
  • #5
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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)??
 
  • #6
Stephen Tashi
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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.
 
  • #7
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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 thats what she answered
 
  • #8
Stephen Tashi
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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.
 
  • #9
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so what would be the distribution of Y then?
 
  • #10
Stephen Tashi
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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 [tex] \phi [/tex]. 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.
 
  • #11
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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.
 
  • #12
Stephen Tashi
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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.
 

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