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  • 0 posts edited
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  • 26 votes cast
Nov
28
accepted How to integrate $\int_{-\infty}^\infty e^{\frac{-(x-y)^2}{2}} dx$
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
asked How to integrate $\int_{-\infty}^\infty e^{\frac{-(x-y)^2}{2}} dx$
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
accepted How is $e^x$ related to its probability density function $\lambda e^{-\lambda x} (x>0)$?
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
accepted Is it possible to do this Poisson problem in Binomial?
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
asked How is $e^x$ related to its probability density function $\lambda e^{-\lambda x} (x>0)$?
Nov
25
accepted Moment generating function: why is $\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^\frac{-(z-t)^2}{2} dz = 1$
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
25
revised Moment generating function: why is $\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^\frac{-(z-t)^2}{2} dz = 1$
deleted 2 characters in body
Nov
25
revised Moment generating function: why is $\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^\frac{-(z-t)^2}{2} dz = 1$
added 49 characters in body
Nov
25
asked Moment generating function: why is $\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^\frac{-(z-t)^2}{2} dz = 1$