Deriving the Mean and Variance of a Continuous Probability Distribution
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- Опубликовано: 22 авг 2024
- I work through an example of deriving the mean and variance of a continuous probability distribution. I assume a basic knowledge of integral calculus.
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Thank you very much for being clear. My university taught calculating E(X) for Discrete Random Variables, but did not go into E(X) for Continuous Random Variables, which this video explains very well.
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Why is the definition of the E(X) equal to that integral in particular? The pdf f(x) value doesn't actually represent the probability of the x value it corresponds to, so then why is the definition of E(X) almost identical to the definition of E(X) for a discrete random variable? I am having some trouble understanding what I am summing with the integral.
See this video: ruclips.net/video/Vyk8HQOckIE/видео.html
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The video that you commented on previously ("Finding probabilities and percentiles for a continuous probability distribution") works through an example of finding the median (and possibly another percentile). Your problem is very similar to the one in that video, the main difference simply being that the function being integrated is not the same.
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jbstatistics
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Almost got it. I have a confusion here.while you were deriving expected value,1/60(x^5/5)... How did we get 5. I have been stuck here for quite sometime and no one is able to explain so far
That's a basic integration: the integral of x^4 is x^5/5 (+c). In other words, the derivative of x^5/5 is x^4.
Great video, really helpful refresh rusty memory, thank you so much! but just one question, at 5:06, should the integration equation be using f(x) instead of F(x), can you please clarify. Thank you
It is f(x), and that's what I wrote. But it can be a little tough to distinguish f(x) from F(x) in handwriting. Cheers.
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Hello jbstatistics. The limits of your pdf are x greater than or equal to 2 and x less than or equal to 4. Suppose the limits were greater than and less than without the equal, would you still integrate using the exact limits of 2 and 4?
Yes. If X is a continuous random variable and c is any constant, then P(X=c) = 0. (Probabilities are areas under the curve, and the area above any point is 0.) So P(X
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When finding (e^2) why is the power suddenly being raised to 5 from 4
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what should i do if the mean i get is no inside the limits?
I do see some formulas where the integral of f(x) is multiplied by 1/b-a where a and b are the interval. what's the difference?
I really don't know what you're asking. If you rephrase it I might understand.
thank you
In my understanding, what was done here is compute, not derive. I expected the explanation behind the formulas for the mean and variance of the distribution.
Those are the *definitions* of the mean and the variance of a continuous probability distribution. Any explanation would simply be describing why those are meaningful quantities to know. I took the definitions of the mean and variance of a continuous random variable, and, using a series of logical and mathematical steps, figured out what the values are for this particular distribution.
But I want to know how did you find the equation of the expectation of Continuous random variable.
Why this equation is equal to mean or expectation?
That's the mean of a continuous random variable by definition. It makes sense to weight each value by the likelihood of it occurring, and the result is the value that the sample mean tends to as the sample size goes to infinity, but the formula is the formula by definition.
does this means that formula foe the variance is same for discrete and continuous variable?
Yes, in the sense that for any random variable X, Var(X) = E[(X - mu)^2].
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At 4:50 you said F(x), which confused me because I thought you meant the cdf... otherwise great video, but it's definitely the pdf here and not the cdf!
Yes, it's the pdf. I wrote a lower case f, and it looks like a lower case f to me. On the line immediately following, in the integrand I replaced "f(x)" with 1/60 x^3, which is given as f(x) (between 2 and 4) above, so I don't think there can be much confusion for long. In handwriting, it can be tough to distinguish a lower case f from an upper case F sometimes, but my upper case F would never cross the vertical line like that.
If we want to calculate the standard deviation. Then we use the variance and it is √0.266 right?
Yes.
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When I intergrate from 4 to 248/75, I get 0.56853? Why is it not 50% or even really close to 50 percent?
You found the area to the left of the mean. The area to the left of the *median* is 0.50, but the area to the left of the mean can be pretty much anything, depending on the shape of the distribution. If it's a symmetric distribution, then the mean and median are equal, but this distribution is definitely not symmetric.
@@jbstatistics Thank you for the clarification. I am taking an econometrics class at UCSD and This channel is really helpful.
Ive been given an exponential function and i need to find Q1 Q2 and Q3, can you give me a hint? all i have is 2exp-2x for x
Anything about marginal and conditional distributions?
i don't understand at 2:02 how do you get the 1/60 . (x^5/5)
+Killan TRAPSA never mind i understood!! man u are a master!
because x.f(x). so x .x^3 = intagrate x^4=> x^5/5 . i hope i could help you.
Thank you, Sir :))
You are very welcome!
sorry, could you please explain why variance of x = 0,266. ? I tried lots of time but I just got the minus sign before
I'm not sure what you are asking. Finding the variance took several minutes in the video, and I don't see how I could flesh out the explanation more than what is given in the video. The final calculation is correct, and if you're not getting the right answer there I'm not sure where you are going wrong.
jbstatistics sorry for my mistake. I mean how did you calculate 56/5 - (248/75)^2=1496/5625. My result was -216... Idk why 🤔😞
I'm not sure what you are doing wrong. Try this in your calculator: (56/5)-(248/75)^2.
why do u increase the power of x when taking out 1/160 out of the integral? at (1.56)
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+hardeep josen You are welcome!
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Is it same formula with moment about mean?
The mean is the first moment about the origin. The variance is the second moment about the mean.
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Intigrate how calculate in3rd step
How is this derivation?
Standard Deviation?
The standard deviation is by definition the (positive) square root of the variance. So to find the standard deviation we first find the variance then take the square root.
Intuitively, I would expect E(X^2) = Integral(x^2*f(x^2)*dx)....
I get a (very reasonable) comment like this from time to time on a number of different videos. Last week I gave this reply to a similar question:
We could find E(g(X)) by using E(g(X)) = sum g(x)f(g(x)), where the summation is over all possible values of g(x) and f(g(x)) = P(g(X) = g(x)). But the law of the unconscious statistician tells us that E(g(X)) also equals sum g(x)p(x), where the summation is over all possible values of x and p(x) = P(X=x). This saves us from having to find the distribution of g(X).
(In your comment, you'd also need dx^2.)
jbstatistics thanks! I did think of dx^2 when I was writing it out. Your explanation makes sense!
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