user133466
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 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 $\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^\frac{-(z-t)^2}{2} dz = 1$ Nov 21 accepted I don't understand why $X = \sum_{i=1}^n X_i$ and $Y = \sum_{i=1}^n Y_i$ 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 $Y = \sum_{i=1}^n Y_i$ Nov 21 comment I don't understand why $X = \sum_{i=1}^n X_i$ and $Y = \sum_{i=1}^n Y_i$ Nov 21 comment I don't understand why $X = \sum_{i=1}^n X_i$ and $Y = \sum_{i=1}^n Y_i$ I 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? Nov 21 comment I don't understand why $X = \sum_{i=1}^n X_i$ and $Y = \sum_{i=1}^n Y_i$ i've been told that capital X and Y are random variables, I never knew you can obtain the actual value of X and Y by summing $X_k$ and $Y_k$. This summation method to obtain X and Y would not work if X and Y are, say, binomially distributed correct? Nov 21 comment I don't understand why $X = \sum_{i=1}^n X_i$ and $Y = \sum_{i=1}^n Y_i$ i've been told that capital X and Y are random variables, I never knew you can obtain the actual value of X and Y by summing $X_k$ and $Y_k$. This summation method to obtain X and Y would not work if X and Y are, say, binomially distributed correct? Nov 21 asked I don't understand why $X = \sum_{i=1}^n X_i$ and $Y = \sum_{i=1}^n Y_i$ Nov 18 accepted Why is $\operatorname{Var}(X+Y) = \operatorname{Cov}(X,X) + \operatorname{Cov}(X,Y) + \operatorname{Cov}(Y,X) + \operatorname{Cov}(Y,Y)$ Nov 18 asked Why is $\operatorname{Var}(X+Y) = \operatorname{Cov}(X,X) + \operatorname{Cov}(X,Y) + \operatorname{Cov}(Y,X) + \operatorname{Cov}(Y,Y)$ Nov 16 comment Does SSTR (sum of squares for treatments) = SSR (regression sum of squares)? hold on a sec, your statement $SSR=\sum\nolimits_{i=1}^{N}(\hat{Y_{i}}-\bar{Y})^{2}=\sum\nolimits_{j=1}^{q}n_{‌​j}(\bar{Y}_{j\cdot}-\bar{Y})^{2}=SST$. means $SSR = SST$??!!! Nov 15 accepted Does SSTR (sum of squares for treatments) = SSR (regression sum of squares)? Nov 14 revised Does SSTR (sum of squares for treatments) = SSR (regression sum of squares)? edited title Nov 14 revised Does SSTR (sum of squares for treatments) = SSR (regression sum of squares)? added 59 characters in body