What is the expected value for Y in the given limiter?

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In summary, for the given limiter, the expected value of Y is equal to the sum of three separate expected values: E[Y1], E[Y2], and E[Y3]. E[Y1] is equal to -a multiplied by the cumulative distribution function of X evaluated at -a. E[Y2] and E[Y3] are integrals with limits of integration from -a to a and from a to infinity, respectively. E[Y2] cannot be simplified further without knowing the specific form of f(x).
  • #1
magnifik
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For the limiter shown below, find the expected value for Y = g(X)
dh6b6s.png


attempt at solution:
E[Y] = ∫g(x)f(x)dx, where f(x) is the probability density function with respect to x
so...
E[Y] = E[Y1] + E[Y2] + E[Y3]
E[Y1] = ∫-af(x)dx where the limits of integration are from -∞ to -a
so E[Y1] = -aFx(-a), where Fx is the cumulative distribution function
E[Y2] = ∫xf(x)dx where the limits of integration are -a to a
this is the part I'm having trouble with
E[Y3] = ∫af(x)dx where limits of integration are a to ∞
so E[Y3] = a(1-Fx(a))

i'm having trouble simplifying E[Y2] into one expression because of the limits of integration. i know it would be just E[Y2] if the limits were -∞ to ∞, but that is not the case here
 
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  • #2
I think that when the domain of integration is limited over a certain set A, the expected value becomes the conditional expected value of X given that X lies on A, times the probability that X lies on A.
 
  • #3
magnifik said:
For the limiter shown below, find the expected value for Y = g(X)
dh6b6s.png


attempt at solution:
E[Y] = ∫g(x)f(x)dx, where f(x) is the probability density function with respect to x
so...
E[Y] = E[Y1] + E[Y2] + E[Y3]
E[Y1] = ∫-af(x)dx where the limits of integration are from -∞ to -a
so E[Y1] = -aFx(-a), where Fx is the cumulative distribution function
E[Y2] = ∫xf(x)dx where the limits of integration are -a to a
this is the part I'm having trouble with
E[Y3] = ∫af(x)dx where limits of integration are a to ∞
so E[Y3] = a(1-Fx(a))

i'm having trouble simplifying E[Y2] into one expression because of the limits of integration. i know it would be just E[Y2] if the limits were -∞ to ∞, but that is not the case here

Unless you have a specific form for f(x) you have gone as far as you can: E(Y2) does not simplify further in any useful way. By the way: I would argue against your notation E(Y1), E(Y2), etc. You do not have three separate random variables Y1, Y2 and Y3; you just have three separate "pieces" of the same random variable Y.

RGV
 
  • #4
hmm..ok
 

FAQ: What is the expected value for Y in the given limiter?

What exactly is a limiter?

A limiter is a device or algorithm used in signal processing to restrict the amplitude of a signal to a predetermined level. It is often used to prevent distorted or clipped signals from causing damage to audio equipment.

How is the expected value of a limiter calculated?

The expected value of a limiter is calculated by taking the average of all the possible outcomes of the limiter. This is done by multiplying each outcome by its corresponding probability and summing them all together.

What factors can affect the expected value of a limiter?

The expected value of a limiter can be affected by the threshold level set for the limiter, the shape of the input signal, and the type of limiter used. Other factors such as noise and distortion can also affect the expected value.

How does the expected value of a limiter relate to the actual output?

The expected value of a limiter provides a theoretical prediction of the average output of the limiter. The actual output may vary from the expected value due to random fluctuations and external factors.

Can the expected value of a limiter be used to accurately predict its performance?

The expected value of a limiter can provide a general understanding of its performance, but it does not guarantee the exact outcome. Other factors such as the shape and characteristics of the input signal can also greatly affect the performance of a limiter.

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