user133466
Reputation
486
Top tag
Next privilege 500 Rep.
Access review queues
 Dec16 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 Nov26 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 $z^2$ Nov26 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? Nov26 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? Nov26 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$(\lambda)$ = $e^\lambda$? Nov26 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!! Nov25 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 Nov25 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? Nov25 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! Nov25 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 $e^{a+b}$ Nov25 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 $\mu = t$... Nov25 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? Nov21 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!! Nov21 comment I don't understand why $X = \sum_{i=1}^n X_i$ and $Y = \sum_{i=1}^n Y_i$ Nov21 comment I don't understand why $X = \sum_{i=1}^n X_i$ and $Y = \sum_{i=1}^n Y_i$ Nov21 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 $X=X_1+...+X_n$ ... Nov21 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 $\sum_{i=1}^n X_i$? Nov21 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 $\sum_{i=1}^n X_i$? Nov21 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 Nov21 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?