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seen Sep 13 '13 at 23:49

Dec
16
comment Is the skill to learn new math by reading textbook alone (no lectures) required when one becomes a PhD student?
in college, instructors always recommended students reading the section that's going to be taught before going to the lecture to increase the effectiveness of the lecture. But I often find myself spending more time reading the textbook on materials that the professor might not have covered. I'm not sure I see how reading prior going to the lecture is beneficial to a student... can you help to elaborate on that a little? Thanks
Nov
26
comment How to integrate $\int_{-\infty}^\infty e^{\frac{-(x-y)^2}{2}} dx$
sorry, im still stuck... i know the integral of $e^x$ is $e^x$ but im not sure how to handle the $z^2$
Nov
26
comment How is $e^x$ related to its probability density function $\lambda e^{-\lambda x} (x>0)$?
i got it, i believe you meant to write "that the probability density function of the random variable $X$ is $\lambda e^{−λx}$ [instead of $ e^{−λx}$] when $x>0$" correct?
Nov
26
comment How is $e^x$ related to its probability density function $\lambda e^{-\lambda x} (x>0)$?
so $X \sim \text{Exp}(\lambda)$ = $\lambda e^{-\lambda x}$ if $x\ge0$ like this?
Nov
26
comment How is $e^x$ related to its probability density function $\lambda e^{-\lambda x} (x>0)$?
so Exp($\lambda$) = $\lambda e^{-\lambda x}$ if $x\ge 0$ where as exp$(\lambda)$ = $e^\lambda$?
Nov
26
comment How is $e^x$ related to its probability density function $\lambda e^{-\lambda x} (x>0)$?
@DilipSarwate Inquest I'm asking that because i'd have to deal with $exp(x)$ and the pdf of an exponential function in this problem. (math.stackexchange.com/questions/244564/…) Others have mentioned that I'm getting confusing the density of an exponential function. If you could shine some light on this question? really appreciated!!
Nov
25
comment Moment generating function: why is $\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^\frac{-(z-t)^2}{2} dz = 1$
i thought the standard form is$\int_{-\infty}^{\infty}(\frac{1}{\sqrt{2\pi}}e^{\frac{-(x-\mu)^2}{2\sigma^2}}‌​)dx = 1$ which in our case is $\int_{-\infty}^{\infty}(\frac{1}{\sqrt{2\pi}}e^{\frac{-(x-t)^2}{2w^2}})dx = 1$ notice: no $\sqrt{w}$ at the denominator of 2$\pi$ and $2w^2$ at exponent's denominator
Nov
25
comment Moment generating function: why is $\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^\frac{-(z-t)^2}{2} dz = 1$
does this mean $\sigma$ can be, say, $w$ and we'll still get $1$ for the integral?
Nov
25
comment Moment generating function: why is $\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^\frac{-(z-t)^2}{2} dz = 1$
this is easier to understand. thank you!
Nov
25
comment Moment generating function: why is $\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^\frac{-(z-t)^2}{2} dz = 1$
@TenaliRaman does $exp(a+b)$ mean $e^{a+b}$
Nov
25
comment Moment generating function: why is $\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^\frac{-(z-t)^2}{2} dz = 1$
err... it's just no where does it say $mu = t$! the denominator fits because the problem stated that $\sigma = 1$ but its like asking me to take a leap when we replace $\mu = t$...
Nov
25
comment Moment generating function: why is $\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^\frac{-(z-t)^2}{2} dz = 1$
what does R.V. stand for?
Nov
21
comment I don't understand why $X = \sum_{i=1}^n X_i$ and $Y = \sum_{i=1}^n Y_i$
still, thank you for your help!!
Nov
21
comment I don't understand why $X = \sum_{i=1}^n X_i$ and $Y = \sum_{i=1}^n Y_i$
let us continue this discussion in chat
Nov
21
comment I don't understand why $X = \sum_{i=1}^n X_i$ and $Y = \sum_{i=1}^n Y_i$
let us continue this discussion in chat
Nov
21
comment I don't understand why $X = \sum_{i=1}^n X_i$ and $Y = \sum_{i=1}^n Y_i$
I don't see how $E(X) = E(X_1+...+X_n)$ is related to $X=X_1+...+X_n$ ...
Nov
21
comment I don't understand why $X = \sum_{i=1}^n X_i$ and $Y = \sum_{i=1}^n Y_i$
Does that mean if we have $X_1,X_2,...X_n$ are independent Poisson, $X$ would still equal to $\sum_{i=1}^n X_i$?
Nov
21
comment I don't understand why $X = \sum_{i=1}^n X_i$ and $Y = \sum_{i=1}^n Y_i$
I understand $E(X)=E(X_1+...+X_n)$ but I did not know $X = X_1 + ... +X_n$. Does that mean if we have $X_1,X_2,...X_n$ are independent Poisson, $X$ would still equal to $\sum_{i=1}^n X_i$?
Nov
21
comment I don't understand why $X = \sum_{i=1}^n X_i$ and $Y = \sum_{i=1}^n Y_i$
looks like I need to look into indicator random variable, the explanation on [wikipedia][2] was not easy to understand. Do you know if there's a better explanation for the usage of indicator random variable? [2]: en.wikipedia.org/wiki/Indicator_function
Nov
21
comment I don't understand why $X = \sum_{i=1}^n X_i$ and $Y = \sum_{i=1}^n Y_i$
say we have $X$ ~ $Bin(n,p)$ what would big $X$ equal to in this case?