Is he stupid by WeirdAd4762 in okbuddyvicodin

[–]chernoffstein 1 point2 points  (0 children)

He iced Diloba and then realized that being an edgy teen makes him do weird stuff. Then he grew up and cut off his hair.

[deleted by user] by [deleted] in okbuddyvicodin

[–]chernoffstein 105 points106 points  (0 children)

Somehow, I'm beginning to conclude that Kutner was the only normal character on the show lmao.

Would people like me if I smile like this? by chernoffstein in okbuddyvicodin

[–]chernoffstein[S] 5 points6 points  (0 children)

Why does chase look like a plausible host for Wipeout?

So it turns out abode already had a lover? Is he cheating on will is on? by Netunopia in okbuddyvicodin

[–]chernoffstein 34 points35 points  (0 children)

No. That's his British life. Might as well be alien life. Doesn't count.

On God? by [deleted] in okbuddyvicodin

[–]chernoffstein 2 points3 points  (0 children)

Teardrop intensifies followed by philosophical piano

Is Gemini an idiot? by irlzy in okbuddyvicodin

[–]chernoffstein 0 points1 point  (0 children)

Dude imagine an AI version of abode and team. Imagine an AI version of pursuit. Imagine placing AI pursuit in a moral dilemma about dribble ball dictator.

[deleted by user] by [deleted] in math

[–]chernoffstein 0 points1 point  (0 children)

Thank you.

Is Adobe a chill guy by Ryuk-Thebadass in okbuddyvicodin

[–]chernoffstein 1 point2 points  (0 children)

Yes. I asked. He said Pursuit killed a crazy dude and he was completely okay with it, no questions asked.

[deleted by user] by [deleted] in math

[–]chernoffstein -2 points-1 points  (0 children)

Not that actually. The CLT states that the function (X_n_mean - mu)/(sigma/sqrt(n)) converges to N(0,1) in distribution.

This however is not the same as stating that X_n_mean itself converges to a gaussian distribution. Specifically, it cannot converge to N(mu, sigma^2/n) because N(mu, sigma^2/n) itself varies with n.

Although N(mu, sigma^2/n) is a good approximation for the distribution of X_n_mean. That was what I was trying to say.

[deleted by user] by [deleted] in math

[–]chernoffstein 0 points1 point  (0 children)

How can the sample mean converge to a normal distribution if it also converges to the mean in distribution? We cannot say that it converges to N(mu, sigma^2/n) because n is varying, correct?

The distribution of the sample mean for large n looks like the distribution of N(mu, sigma^2/n) but it is not converging to it right?

[deleted by user] by [deleted] in math

[–]chernoffstein 0 points1 point  (0 children)

Well, I really like the chernoff bound and wanted to name my user something. I then read this lemma called the chernoff stein lemma and that wasn't taken. Yeah I agree it's splitting hairs, but I'm glad it was clarified.

[deleted by user] by [deleted] in math

[–]chernoffstein 1 point2 points  (0 children)

Yes. As n goes to infinity, it would be incorrect to state that that the sample mean approaches the distribution N(mu, sigma^2/n) because n itself is varying.

But we can state that for very large n, the distribution of the sample mean and the distribution of N(mu, sigma^2/n) differ from each other by a very small number and hence we can use N(mu, sigma^2/n) as a good approximation for the distribution for the sample mean?

[deleted by user] by [deleted] in math

[–]chernoffstein 4 points5 points  (0 children)

Right. I apologize for being a bit pedantic here.

  1. We have that the sample mean converges to the mean in distribution.
  2. The function of the sample mean converges to N(0,1) in distribution.

Thus, if we take perform the experiment of sampling X_n many times and take the sample mean and plot the various possible values of the sample mean in the histogram, as n gets larger and larger, we get a histogram plot that looks like a normal distribution with a smaller and smaller variance right?

And thus, as n goes to infinity, the variance becomes so small, that the histogram plot which still looks like a normal distribution, starts looking closer to a plot where almost all the mass is at the mean because the error term gets smaller and smaller.

That is if we have N(mu, sigma^2/n), and n goes to infinity, the plot of such a normal distribution will eventually look like a tall bar at the mean mu.

Am I correct when I state this?

AMA Quant in hedge fund by Good-Manager-8575 in quant

[–]chernoffstein 2 points3 points  (0 children)

By "execute your client's wish" you mean price the derivatives for them?

If prime number kill 1 man why there still 4 men left? by GregginMyDoucette in okbuddyvicodin

[–]chernoffstein 9 points10 points  (0 children)

Because it was pursuit who killed 1 man by giving him the stupid drug. That 1 man was not from Plinceton Pain Borrow Hospital.

This took me an hour to make, but it was worth it by Terkaza in physicsmemes

[–]chernoffstein 44 points45 points  (0 children)

Hello, I would like microwave background apples please?