For more solved question on the Joint Probability Density function click on the link below: drive.google.com/file/d/1sEylHsFaX6aHTWjiCKaBqwaLcnfG0A7X/view?usp=sharing
For this video I was using Smooth Draw, and Wacom Bamboo Splash Pen Tablet. Though recently I upgraded my tablet to the XP-PEN Artist12 11.6 Inch FHD Drawing Monitor - I really like both of these for screen writing :-)
could you solve the question by factoring the joint distribution into x(2-y) which shows the two variables are independent. Then by knowing that the marginal density of X is proportional to x you find the the constant "c" which satisfies c * integral (x) from 0 to 1 =1. Then finally you integrate x from 0 to 1/2 and then multiply by to get the answer. I ask because that is what I did.
Let X and Y be independent random variables and both of them are uniformly distributed in [0, 1]. If the smaller (of the two) is less than 1/4, then what is the conditional probability that the larger is greater than 3/4? please solve this
Since X and Y are independent, the the probability that X is less than 1/4 does not depend on Y being greater than 3/4.... In other words: Pr(X < 1/4 | Y > 3/4) = Pr(X < 1/4) since X is independent of Y. To find Pr(X < 1/4) use the uniform distribution and integrate from 0 to 1/4... Let me know if this makes sense, or if you have any follow-up questions
What if Y=1 and not less than. I am assuming you would still have to find the marginal distribution but I don't think you would then take the integral. What would you do instead? Or would you take the integral?
This is a good question. Thank you Jaime for posting. We would start by finding the conditional distribution, fx|y = fxy / fy We found in the video that fy = 1 - 1/2 y, 0
Hi Sameer! You are thinking about X as if it were a discrete number. X is not discrete here. We do not need to reduce 1 to 0.99 because for continuous random variables the probability that X is equal to 1 is zero. Therefore, if I want to know the probability that X is less than 1, I would integrate between 0 and 1 (not between 0 and 0.99). Thanks for the post! if you have any follow-up questions, please let me know
This taught me in the first two minutes what my graduate-level professor was unable to teach over two days. Thanks!
I like your style. You and Mancinelli are the best. Thank you.
For more solved question on the Joint Probability Density function click on the link below:
drive.google.com/file/d/1sEylHsFaX6aHTWjiCKaBqwaLcnfG0A7X/view?usp=sharing
What a big help! You explained it very well. Thank you so much :)
Thank you so so much ! Ugh. Been struggling with this topic 😣
Thanks!
Best explanation
Thanks for the Video
Nice Video. Between which device you are using for writing on screen
For this video I was using Smooth Draw, and Wacom Bamboo Splash Pen Tablet. Though recently I upgraded my tablet to the XP-PEN Artist12 11.6 Inch FHD Drawing Monitor - I really like both of these for screen writing :-)
Thanks
thank you so much
could you solve the question by factoring the joint distribution into x(2-y) which shows the two variables are independent. Then by knowing that the marginal density of X is proportional to x you find the the constant "c" which satisfies c * integral (x) from 0 to 1 =1. Then finally you integrate x from 0 to 1/2 and then multiply by to get the answer. I ask because that is what I did.
Edit of last 2 sentences:
Then finally you integrate x from 0 to 1/2 and then multiply by "c" get the answer. I ask because that is what I did
You're amazing
Let X and Y be independent random variables and both of them
are uniformly distributed in [0, 1]. If the smaller (of the two) is less than 1/4, then what is the
conditional probability that the larger is greater than 3/4? please solve this
Since X and Y are independent, the the probability that X is less than 1/4 does not depend on Y being greater than 3/4....
In other words:
Pr(X < 1/4 | Y > 3/4) = Pr(X < 1/4) since X is independent of Y.
To find Pr(X < 1/4) use the uniform distribution and integrate from 0 to 1/4...
Let me know if this makes sense, or if you have any follow-up questions
What if Y=1 and not less than. I am assuming you would still have to find the marginal distribution but I don't think you would then take the integral. What would you do instead? Or would you take the integral?
This is a good question. Thank you Jaime for posting. We would start by finding the conditional distribution,
fx|y = fxy / fy
We found in the video that
fy = 1 - 1/2 y, 0
@@Stats4Everyone If it is X
in case pr(X
Hi Sameer! You are thinking about X as if it were a discrete number. X is not discrete here. We do not need to reduce 1 to 0.99 because for continuous random variables the probability that X is equal to 1 is zero. Therefore, if I want to know the probability that X is less than 1, I would integrate between 0 and 1 (not between 0 and 0.99). Thanks for the post! if you have any follow-up questions, please let me know
can u please help me how to do for exponential function with x and y
Can you clarify this question. What does the joint pdf look like? Also, what is the probability you are looking to solve?