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 Nov26 asked How to integrate $\int_{-\infty}^\infty e^{\frac{-(x-y)^2}{2}} dx$ 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 accepted How is $e^x$ related to its probability density function $\lambda e^{-\lambda x} (x>0)$? 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 accepted Is it possible to do this Poisson problem in Binomial? 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 asked How is $e^x$ related to its probability density function $\lambda e^{-\lambda x} (x>0)$? Nov25 accepted Moment generating function: why is $\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^\frac{-(z-t)^2}{2} dz = 1$ 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? Nov25 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 Nov25 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 Nov25 asked Moment generating function: why is $\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^\frac{-(z-t)^2}{2} dz = 1$ Nov21 accepted 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 $Y = \sum_{i=1}^n Y_i$ still, thank you for your help!!