Probability density function, cumulative function.

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SUMMARY

The discussion centers on a mixed random variable defined by the probability density function \( p_{\xi}(v) = 2.25u(v+2)e^{-3(v+2)} + k\delta(v-2) \), where \( u \) is the Heaviside step function. The constant \( k \) is determined to be 0.25, indicating a non-zero probability at \( v = 2 \) due to the delta function. The cumulative distribution function \( F(x) \) is derived as \( F(x) = \int_{-\infty}^{x} p(v) \, dv \), which is not continuous at \( x = 2.

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  • Understanding of probability density functions (PDFs)
  • Knowledge of the Heaviside step function
  • Familiarity with delta functions in probability theory
  • Ability to compute integrals for cumulative distribution functions (CDFs)
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  • Study the properties of mixed random variables
  • Learn about the application of delta functions in probability distributions
  • Explore the derivation of cumulative distribution functions from probability density functions
  • Investigate examples of lifetime distributions in reliability engineering
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Bassalisk
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Homework Statement


Random voltage is defined with its probability density function:
[tex]p_{\xi}(v)=2,25u(v+2)e^{-3(v+2)}+k\delta (v-2)[/tex]

u-Heaviside step function

a) Find constant k.
b) What is the probability of a random variable to take value of 2.
c) Find the cumulative distribution function.

Homework Equations


[itex]\int_{-\infty}^{+\infty}p_{\xi}(x)dx=1[/itex]

The Attempt at a Solution


I got that the constant k is 0,25. (if somebody would check that for me)

Now for the b), I am not sure. I know that when you have continuous distribution, probability that the function has 1 specific value is 0.

But this delta function at that point is confusing me. Does it change things? Should the probability be 0.25?

And I have troubles finding cumulative distribution. Help me out here.
 
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Bassalisk said:

Homework Statement


Random voltage is defined with its probability density function:
[tex]p_{\xi}(v)=2,25u(v+2)e^{-3(v+2)}+k\delta (v-2)[/tex]

u-Heaviside step function

a) Find constant k.
b) What is the probability of a random variable to take value of 2.
c) Find the cumulative distribution function.

Homework Equations


[itex]\int_{-\infty}^{+\infty}p_{\xi}(x)dx=1[/itex]

The Attempt at a Solution


I got that the constant k is 0,25. (if somebody would check that for me)

Now for the b), I am not sure. I know that when you have continuous distribution, probability that the function has 1 specific value is 0.

But this delta function at that point is confusing me. Does it change things? Should the probability be 0.25?

And I have troubles finding cumulative distribution. Help me out here.

This is an example of a "mixed" random variable that has both a continuous and discrete component. Other (perhaps more familiar) examples are lifetime distributions of products, which may have a nonzero probability at lifetime = 0, because the product may be non-functional right out of the box, but if it is functional when new, then its remaining life has a continuous distribution. Anyway, that is the role of the delta function: it describes a point-mass for probability at v = 2. (Are you sure you should have delta(v-2) rather than delta(v+2)? It seems strange to have a point mass in the middle of a continuous part.)

So, in your case, Pr{V = 2} = k, which is not zero.

The cumulative distribution is just F(x) = Pr(V <= x} = integral(p(v), v = -infinity..x), which ought to be easy for you to do. In this case F(x) is not continuous at x = 2; maybe that is what was confusing you.

RGV
 
Ray Vickson said:
This is an example of a "mixed" random variable that has both a continuous and discrete component. Other (perhaps more familiar) examples are lifetime distributions of products, which may have a nonzero probability at lifetime = 0, because the product may be non-functional right out of the box, but if it is functional when new, then its remaining life has a continuous distribution. Anyway, that is the role of the delta function: it describes a point-mass for probability at v = 2. (Are you sure you should have delta(v-2) rather than delta(v+2)? It seems strange to have a point mass in the middle of a continuous part.)

So, in your case, Pr{V = 2} = k, which is not zero.

The cumulative distribution is just F(x) = Pr(V <= x} = integral(p(v), v = -infinity..x), which ought to be easy for you to do. In this case F(x) is not continuous at x = 2; maybe that is what was confusing you.

RGV

Thank you for your reply. I am sure that the δ is at 2. We usually have that, unusual things :).
 

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