If we have $f(x,y)=f(\overline{x}\pm\sigma_x,\overline{y}\pm\sigma_y)$ how is then $$\sigma_f=\sqrt{\left(\frac{\partial{f}}{\partial{x}}\right)^2\sigma_x^2+\left(\frac{\partial{f}}{\partial{y}}\right)^2\sigma_y^2}~~~?$$   I have looked at  Wikipedia page but I don't quite understand this approach since I'm not familiar with approximating functions with Taylor series. I have also found that in  video it is said that that way of calculating the standard deviation for multivariate functions somehow gives us a average standard deviation but I don't understand why/how.
1721
I'm in doubt with one thing: let's imagine that we have $n+1$ quantities, $n$ of them being directly measured, and the other one being related to the first $n$ by a function $f : \mathbb{R}^n \to \mathbb{R}$. If the $x_i$ is one of the quantities being directly measured and if $\Delta x_i$ is it's uncertainty, then I've learned that the uncertainty of the quantity $y = f(x_1, \dots x_n)$ is given by:  $$\Delta y=\sum_{i=1}^{n}{\frac{\partial f}{\partial x_i}\Delta x_i}$$  My problem in understanding this is: well, if $f$ is such a function, it's derivative is the linear transformation $df : \mathbb{R}^n \to \mathbb{R}$ given by:  $$df=\sum_{i=1}^{n}\frac{\partial f}{\partial x_i}dx_i$$  Where $dx_i : \mathbb{R}^n \to \mathbb{R}$ is given by $dx_i(a_1 \dots a_n) = a_i$. Hence, what we are saying, is that $\Delta y = df ( \Delta x_1, \dots \Delta x_n)$, in other words, we are saying that the uncertainty of $y$ is given by a linear function of the vector of uncertainties of the quantities $x_i$. Why should that be true? I mean, what's the reasoning behind this? I really didn't get the point, my teacher just gave the equation without justifying or anything like that.  Any help and reference is appreciated.
105744
Let $X, Y$ be independent standard normal random variables.  We already know that $X+Y$ is normal with mean 0 and variance 2. However, I am trying to prove this result in a slightly different way than usual, i.e., using convolutions.  We can address the problem in the following way: the sum of the two variables has a density $h(x)$ given by the convolution product of the densities of the two variables, say $f$ and $g$, so that   $$ h(x) = \int_{-\infty}^{\infty} f(x-y)g(y)dy. $$  However,  $$ f(y) = g(y) = \frac{1}{\sqrt {2\pi}}\exp(-y^2/2). $$  Hence now the integral becomes:  $$ h(x) =\frac{1}{2\pi}\int_{-\infty}^{\infty}\exp(-(x-y)^2/2)\exp(-y^2/2)dy. $$  However, I am puzzled here. How should we go on from here? I cannot see how to transform this integral into the integral of a normal r.v of type $N(0,2)$. Thanks in advance for your help.
96591
Given that a random variable $X$ is a normally distributed variable with mean $\mu$ and variance $\sigma^2$. Why exactly is:  $$\mathbb{E}[|X|] = \mathbb{E}[X\mid X&gt;0]P(X&gt;0) - \mathbb{E}[X\mid X&lt;0]P(X&lt;0)$$  Because we have that:  $$\mathbb{E}[X\mid X&gt;0] = \frac{\int_0^\infty xf_X(x)\,dx}{P(X&gt;0)}$$  Therefore my definition immediately above must be incorrect?  Because if it is not we have that:  $$\mathbb{E}[|X|] = \int_0^\infty xf_X(x)\,dx - \int_{-\infty}^0 xf_X(x)\,dx$$  and where does the minus there come from anyway, and why?
107688
Given the mean &amp; variance of a dataset of linear measurements X ... N~(14,2), I want to find the mean of X^3.    So by taking mean as being expected value, &amp; rearranging: -  $$Var(X)=E[X^2]−(E[X])^2$$  I get   $$ E[X^2] = Var(X) + (E[X])^2$$  $$E[X^2] = SD^2 + (E[X])^2$$  How can I go from this to using an exponent of 3 to solve the original problem?
123566
Given that $a$~$\mathcal{N}(\mu, \Sigma)$ and $\mu$~$\mathcal{N}(0, \sigma^2 I)$, Is $p(a|\Sigma, \sigma)=\int p(a|\mu, \Sigma)p(\mu|\sigma)d\mu$ again a Normal distribution? And how to estimate the mean and variance respectively? Thanks a lot!
116864
According to Wikipedia, the standard deviation of a sample mean is calculated as follows $$\frac{\sigma}{\sqrt{n}}$$  Why is that? Why do we need to divide the standard deviation of the population by the square root of $n$ (which should I think be the size of the sample)? Why should that make sense?
93533
I have a function $y=sin(x)$  I sample the points $ x_i=\{0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1\}$ and $$ y_i=sin(x_i) + \epsilon \  \ \epsilon \sim iid\ Normal$$  Now given only the set $\{x_i,y_i\}$ I want to uncover the underlying functional form $f(x)$ given as  $$y=f(x)+\epsilon \ \ \ \ \epsilon \sim iid\ Normal  $$  My question is : ARE $y_i$'s independent ?  My intuition is that $y_1$ and $y_2$ are not independent. This is because given a new ${x}$ say ${x_\star}$ we can use $y_i$'s to learn some information about ${y_\star}$.   Does the fact that the input points come from the same underlying function make them somehow correlated or posses commonalities ? and if not why do most methods (ex: Non-parametric splines, Gaussian process) use the $y_i$'s to get information and learn ${y_\star}$ ? Isn't the fact that if samples are independent then they can't learn new information from each other ?
50765
I have a random variable $X(a) = \log(a)$ where a is normal distributed $\mathcal N(\mu,\sigma^2)$. What can I say about $E(X)$ and $Var(X)$? An approximation would be helpful too.
60272
Suppose I have 2 sets:  Set A: number of items $n= 10$, $\mu = 2.4$ , $\sigma = 0.8$  Set B: number of items $n= 5$, $\mu = 2$, $\sigma = 1.2$  I can find the combined mean ($\mu$) easily, but how am I supposed to find the combined standard deviation?
83880
I have the mean and median values for a sample drawn from a lognormal distribution. Note that this is not the mean and median of the logs of the variable, though I can of course calculate the logs of the mean and median. Is there a closed form solution for μ and σ from this information? If there is only a numeric solution, could you tell me how to find it, ideally with R?  I note that this question has been answered for deriving μ and σ from the sample mean and the sample variance, here:   However, I do not have the sample variance, only the mean and median.  If there is no closed-form or straightforward numeric solution, I'd like to know if using the logs of the sample mean and median, or some transform of them, will provide a reasonable answer for a large sample (in the hundreds of millions).
771
Let $X_1, X_2, \ldots X_n$ be random variables with mean $\mu$ and variance $\sigma^2$. We define the sample mean $\bar{X} = \sum_{i = 1}^n X_i$.  I am having trouble understanding why the variance of $\bar{X}$ is $\sigma^2 / n$.  Here is what I worked out thus far:  $$\begin{gather} \sigma_\bar{X}^2 = E((\bar{X} - \mu)^2) = E(\bar{X}^2 - 2\bar{X}\mu + \mu^2) = E(\bar{X}^2) - 2\mu E(\bar{X}) + E(\mu^2) = E(\bar{X}^2) - \mu. \end{gather}$$  However, I'm not sure what to do with the $E(\bar{X}^2)$.
59871
I've been given a problem where I have  $$ X \sim \mathcal{N}(\mu = 2, \sigma^2 = 9) $$  $$ Y \sim \mathcal{N}(\mu = 3, \sigma^2 = 4) $$  Their correlation is $ \rho_{XY} = 0.6 $. First I am asked for the mean and standard distribution of X+Y, which I found as  $$ \mu_{X+Y} = \mu_X + \mu_Y = 2 + 3 = 5 $$  $$ \sigma_{X+Y} = \sqrt{\sigma_{X}^2 + \sigma_{Y}^2 + 2 \rho_{XY} \sigma_x \sigma_y} $$  However, I am also asked what the distribution of $ X + Y $ is, what if $X$ and $Y$ are jointly normally distributed, and what if $X$ and $Y$ are not jointly normally distributed. I am unsure how I would go about determining what the distribution of $X + Y$ is, though I assume that    If $X$ and $Y$ are jointly normally distributed then $X + Y$ is normally distributed If $X$ and $Y$ are not jointly normally distributed then $X + Y$ is not normally distributed   How do I determine what the distribution of $X + Y$ is when I am not told whether they are jointly distributed?    Related  I read the following post, but wasn't sure how to apply what was said to my current problem, if I could at all.
120056
[] ** calculate the mean and standard deviation of each investment, assume that both returns come from independent samples, are both normally distributed and that the (true but unknown) variances are the same,, Can someone could check my  calculations for  mean and standard deviation? I did calculations but not sure what to do with negative values of distances(If if am correct Investment A MEAN= 78.6875 when it comes to calculating standard deviation i have to power difference by 2 ?  **
15873
Suppose there is a very big (infinite?) population of normally distributed values with unknown mean and variance.  Suppose also that we have a sample, S, of n values from the entire population. We can calculate mean and standard deviation for this sample (we use n-1 for stdev calculation).  The first and most important question is how is stdev(S) related to the standard deviation of the entire population?  An illustration for this issue is the second question:  Suppose we have an additional number, x, and we would like to test whether it is an vis-a-vis the general population. My intuitive approach is to calculate Z as follows:  $Z = \frac{x - mean(S)}{stdev(S)}$  and then test it against standard distribution if n>30 or against t-distribution if n&lt;30.  However, this approach doesn't account for n, the size of the sample. What is the right way to solve this question provided there is only single sample S?
8182
I need to find the distribution of the random variable  $$Y=\sum_{i=1}^{n}(X_i)^2$$ where $X_i\sim{\cal{N}}(\mu_i,\sigma^2_i)$ and all $X_i$s are independent. I know that it is possible to first find the product of all moment generating functions for $X_i$s, and then transform back to obtain $Y$'s distribution. However, I wonder whether there is a general form for $Y$ like the Gaussian case: we know the sum of independent Gaussian is still a Gaussian, and thus we only need to know the summed mean and summed variance.   How about all $\sigma^2_i=\sigma^2$? Will this condition make a general solution?
3975
Suppose $X$ is a random variable with pdf $f_X(x)$. Then the random variable $Y=X^2$ has the pdf   $$f_Y(y)=\begin{cases}\frac{1}{2\sqrt{y}}\left(f_X(\sqrt{y})+f_X(-\sqrt{y})\right) &amp; y \ge 0 \\ 0 &amp; y \lt 0\end{cases}$$   I understand the calculus behind this. But I'm trying to think of a way to explain it to someone who doesn't know calculus. In particular, I'm trying to explain why the factor $\frac{1}{\sqrt{y}}$ appears out front. I'll take a stab at it:  Suppose $X$ has a Gaussian distribution. Almost all the weight of its pdf is between the values, say, $-3$ and $3.$ But that maps to 0 to 9 for $Y$. So, the heavy weight in the pdf for $X$ has been extended across a wider range of values in the transformation to $Y$. Thus, for $f_Y(y)$ to be a true pdf the extra heavy weight must be downweighted by the multiplicative factor $\frac{1}{\sqrt{y}}$  How does that sound?  If anyone can provide a better explanation of their own or link to one in a document or textbook I'd greatly appreciate it. I find this variable transformation example in several intro mathematical probability/stats books. But I never find an intuitive explanation with it :(
53462
What is the variance of the sample variance? In other words I am looking for $\mathrm{Var}(S^2)$.  I have started by expanding out $\mathrm{Var}(S^2)$ into $E(S^4) - [E(S^2)]^2$  I know that $[E(S^2)]^2$ is $\sigma$ to the power of 4. And that is as far as I got.