Exponential Distribution and median

In summary, the conversation discusses finding the (100p)th percentile of a distribution with parameter \lambda, starting with the exponential distribution. The formula for the exponential distribution is given and the conversation includes attempts at finding the median using the cumulative distribution function (cdf). The conversation also clarifies the difference between probability density function and cumulative distribution function. The correct antiderivative for the cdf is discussed and the conversation concludes with a reminder to solve for the value of x to find the 100pth percentile.
  • #1
exitwound
292
1

Homework Statement



IF X has an exponential distribution with parameter [itex]\lambda[/itex], derive a general expression for the (100p)th percentile of the distribution. Then specialize to obtain the median.

Homework Equations



[tex]X.exp(\lambda)=\lambda e^{-\lambda x} for x>0[/tex]

[tex]p=F(\eta (p)) = \int_{-\infty}^{\eta (p)} f(y)dy[/tex]

The Attempt at a Solution


First, can I verify that this antiderivative is correct?
Since F'(X) = f(X) [tex]F(X)=-e^{-\lambda x}[/tex]
Finding the 100pth percentile is equivalent to finding the cumulative density function (the antiderivative) from 0-->p correct?

And F(.5) = [itex]\eta[/itex] for the mean...

I'm really confused on how to set this up and what I'm looking for. Any help would be great.
 
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  • #2
Your limits of integration are correct, you are just missing one term in the expression for P(X≤x)


[tex]P(X \leq x) = \int _0 ^x \lambda e^{-\lamda t} dt[/tex]


re-integrate it and you'll see.

Also I think you want to find the median 'm' such that F(m)=1/2 or in the integral form

[tex]\int _0 ^m \lambda e^{-\lamda t} dt =1/2[/tex]
 
  • #3
I don't follow at all. Can you try to explain it to me in english rather than math? I'm not a math major and the language confuses me more than anything else.
 
  • #4
When you write cumulative density function you probably mean cumulative distribution function. As far as I know cumulative density function has no meaning in probability theory and Wikipedia writes:
Cumulative density function is a self-contradictory phrase resulting from confusion between:
- probability density function, and
- cumulative distribution function.
The two words cumulative and density contradict each other.
However from you usage it seems clear that you intend for it to be cumulative distribution function (cdf).

exitwound said:
First, can I verify that this antiderivative is correct?
Since F'(X) = f(X) [tex]F(X)=-e^{-\lambda x}[/tex]
This is the right antiderivative for non-negative numbers however it may not be exactly the one you're looking for (remember there are infinitely many that can be obtained by adding constants). Remember that f(x)=0 for x<0 so the antiderivative is 0 if x < 0.
[tex]G(x) = \begin{cases} -e^{-\lambda x} & 0 \leq x \\ 0 & x < 0 \end{cases}[/tex]
is probably what you meant.

Now as you wrote earlier the cdf is:
[tex]F(x) = \int_{-\infty}^x f(y) dy = G(x) - \lim_{y \to -\infty}G(y)[/tex]
It's not enough for it be an antiderivative of f(y), it must be the right one (given by the equation above).
From that you should be able to compute the correct cdf (and it's not equal to G(x)).

Finding the 100pth percentile is equivalent to finding the cumulative density function (the antiderivative) from 0-->p correct?
I'm not sure exactly what finding the cdf from 0->p means, but I don't think it's the right idea. By definition we say that a value x is the 100pth percentile if F(x)=p. Thus you need to solve this equation for x to find an expression for the 100pth percentile.

And F(.5) = [itex]\eta[/itex] for the mean...
The mean is just a fancy word for the 50th percentile. That is we call m the mean if F(m)=.5
 

What is an exponential distribution?

An exponential distribution is a probability distribution that describes the time between events in a Poisson process, where events occur continuously and independently at a constant average rate. It is often used to model the time between arrivals of customers in a queue or the time between failures of a machine.

How is the exponential distribution related to the median?

The median of an exponential distribution is the value at which the cumulative distribution function (CDF) equals 0.5. This means that there is a 50% chance that a random observation from the distribution will be less than the median. It can also be thought of as the "half-life" of the distribution, where half of the observations occur before the median and half occur after.

What is the formula for calculating the median of an exponential distribution?

The formula for calculating the median of an exponential distribution is: median = (ln 2)/λ, where λ is the rate parameter of the distribution. This formula can also be written as median = 1/λ, since ln 2 ≈ 0.693.

Can the median of an exponential distribution be greater than the mean?

Yes, it is possible for the median of an exponential distribution to be greater than the mean. This occurs when the distribution is "skewed" to the right, meaning that there are a few large values that pull the mean upward while the majority of the data is clustered towards the lower end. In this case, the median may be a more representative measure of central tendency than the mean.

How is the exponential distribution used in real-world applications?

The exponential distribution is used in a variety of real-world applications, such as modeling waiting times in queuing systems, predicting the time between failures of machines, and analyzing the inter-arrival times of customer purchases. It is also commonly used in survival analysis to model the time until an event (such as death) occurs. Additionally, it is often used in financial modeling to estimate the probability of extreme events, such as stock market crashes.

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