The logic here I understand is: if we assume Prob(x>=6) > 7/12, the expectation will greater than 3.5 as the prob(x=1/2/3/4/5) all >0. This contradicts with that E(X)=3.5, and hence Prob(x>=6) must
One intuition is that more than half of a population of N can't have value of more than double the mean mu. If it were possible then you'd have at least N/2 individuals, with value at least 2 * mu. Just from these values you'd already have the mean because ( N / 2 * 2 * mu ) / N = mu, and you've still got another N/2 individuals left that are greater than or equal to zero! It's very important to note this only works for the case where values are positive, the inequality doesn't hold if you can have negative values.
Hi, Markov's inequality provides an upper limit to the probability. As your example demonstrates (although I think the RHS should be 5 rather than 3?) it is not always useful, but nonetheless true. Hope that helps! Thanks, Ben
this is a pretty mechanical explanation, if this is this if we assume this if we if if if, tells us nothing about the intuition behind it. poorly thought example, all you did was just a mindless calculation and you proved it is not but why it is not was never mentioned and what the logic behind it is. p.s: you just gave us markov's inequality but never explained what is the logic of dividing by a and why is the probability of a greater or equal than X equals x equal to the expected value of it and then divided by a, looks like a random game someone made up.
markov inequality always felt like magic to me, thanks for proving the intuition for it :-)
This video deserves many more views. Thanks!
The logic here I understand is: if we assume Prob(x>=6) > 7/12, the expectation will greater than 3.5 as the prob(x=1/2/3/4/5) all >0. This contradicts with that E(X)=3.5, and hence Prob(x>=6) must
Yeah spend some time to understand this, u summed it up very well
One intuition is that more than half of a population of N can't have value of more than double the mean mu. If it were possible then you'd have at least N/2 individuals, with value at least 2 * mu. Just from these values you'd already have the mean because ( N / 2 * 2 * mu ) / N = mu, and you've still got another N/2 individuals left that are greater than or equal to zero! It's very important to note this only works for the case where values are positive, the inequality doesn't hold if you can have negative values.
Hi,
Markov's inequality provides an upper limit to the probability. As your example demonstrates (although I think the RHS should be 5 rather than 3?) it is not always useful, but nonetheless true. Hope that helps!
Thanks,
Ben
Very good video!! Keep up the good job.
This is intution or giving proof by contradiction??
I understand this video much better than the mathematical model of Markov's!
Hi, glad to hear that it helped! All the best, Ben
Thanks!
If the random variable X is uniform between 1 and 4, Markov inequality gives, P(X>0.5)
If that was the case wouldn't the expected value also change based on the probability distribution being used?
this is a pretty mechanical explanation, if this is this if we assume this if we if if if, tells us nothing about the intuition behind it.
poorly thought example, all you did was just a mindless calculation and you proved it is not but why it is not was never mentioned and what the logic behind it is.
p.s: you just gave us markov's inequality but never explained what is the logic of dividing by a and why is the probability of a greater or equal than X equals x equal to the expected value of it and then divided by a, looks like a random game someone made up.
This was actually a good explanation, but you need to understand mathematical expectancy first.
@2423 6585 What useful buried treasure. Thank you!
since when is P(X=6) in rolling a dice 7/12?? its 1/6 what am i missing wtf
It's not the probability itself, it's the upper bound! 1/6 is less than 7/12? Yeah, so Markov's Inequality works :)
Terrible explanation man, terrible explanation
what's bad about it? I found it quite good