I don't understand why $Var(X) = E((X-\mu)^2)$.   It's defined as the "expected value of the square of the deviation of $X$ from the mean" but I don't understand why it couldn't be $E(X-\mu)$ as that seems more intuitive for "deviation from the mean".  Is the purpose of the squaring to make deviations positive? Because if so, why not just $E(|X-\mu|)$ instead? What's the point of the squaring?
329
Let's take the numbers 0-10.  Their mean is 5, and the individual deviations from 5 are -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5 And so the average (magnitude of) deviation from the mean is $30/11 \approx 2.72$.  However, this is not the standard deviation.  The standard deviation is $\sqrt{10} \approx 3.16$.  The first mean-deviation is a simpler and by far more intuitive definition of the "standard-deviation", so I'm sure it's the first definition statisticians worked with.  However, for some reason they decided to adopt the second definition instead.  What is the reasoning behind that decision?
37683
Just wondering if it is possible to find the Expected value of x if it is normally distributed, given that  is below a certain value (for example, below the mean value).
72308
I've been reading over this , trying to make sense of how the mean and variance of a gaussian conditional was derived. I've come to accept that unless I allocate a dozen or so hours to refreshing my linear algebra knowledge, it's out of my reach for the time being. that being said, I'm looking for a conceptual explanation for that these equations represent: $$\mu_{1|2} = \mu_1 + \Sigma_{1,2} * \Sigma^{-1}_{2,2}(x_2 - \mu_2)$$ I read the first as &quot;Take $\mu1$ and augment it by some factor, which is the covariance scaled by the precision (measure of how closely $X_2$ is clustered about $\mu_2$, maybe?) and projected onto the distance of the specific $x_2$ from $mu_2$.&quot; $$\Sigma_{1|2} = \Sigma_{1,1} - \Sigma_{1,2} * \Sigma^{-1}_{2,2} * \Sigma_{1,2}$$ I read the second as, &quot;take the variance about $\mu_1$ and subtract some factor, which is covariance squared scaled by the precision about $x_2$.&quot; In either case, the precision $\Sigma^{-1}_{2,2}$ seems to be playing a really important role. A few questions:  Am I right to treat precision as a measure of how closely observations are clustered about the expectation? Why is the covariance squared in the latter equation? (Is there a geometric interpretation?) So far, I've been treating $\Sigma_{1,2} * \Sigma^{-1}_{2,2}$ as a ratio, (a/b), and so this ratio acts to scale the (second) $\Sigma_{1,2}$, essentially accounting for/damping the effect of the covariance; I don't know if this is valid. Anything else you'd like to add/clarify?
87060
I have a Geometric Distribution, where the stochastic variable $X$ represents the number of failures before the first success.  The distribution function is $P(X=x) = q^x p$ for $x=0,1,2,\ldots$ and $q = 1-p$.  Now, I know the definition of the expected value is: $E[X] = \sum_{i}{x_i p_i}$  So, I proved the expected value of the Geometric Distribution like this:  $E[X]=\sum _{ i=0 }^{ \infty  }{ iP(X=i) } = \sum _{i=0}^{\infty}{i q^i p} = p\sum _{i=0}^{\infty}{i q^i} = pq \sum _{i=0}^{\infty}{iq^{i-1}}$  $\qquad = pq \sum _{i=0}^{\infty}{\frac{d}{dq}q^i} = pq \frac{d}{dq}(\sum _{i=0}^{\infty}{q^i}) = pq \frac{d}{dq}(\frac{1}{1-q})$  $\qquad = pq \frac{1}{(1-q)^2} = \frac{pq}{p^2} = \frac{q}{p}$  So now, I would like to prove that $Var[X] = \frac{q}{p^2}$. I know I have to use a simular trick as above (with the derivation).  $Var[X] = E[X^2] - E[X]^2 = \sum _{i=0}^{\infty}{i^2 q^i p} - (\frac{q}{p})^2 = p \sum _{i=0}^{\infty}{i^2 q^i} - (\frac{q}{p})^2 = pq \sum _{i=0}^{\infty}{i^2 q^{i-1}} - (\frac{q}{p})^2$  $\qquad = pq \sum _{i=0}^{\infty}{\frac{d}{dq}i q^i} - (\frac{q}{p})^2 = pq \frac{d}{dq} \sum _{i=0}^{\infty}{iq^i}-(\frac{q}{p})^2$  Then I'm stuck. How can I get another $q$ out of the sum? Won't it mess up the first derivation?
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?
71663
I just read this very insightful post , where the author stated that the variance of $\hat\beta$ is:  $$\text{var}(\hat\beta) = \sigma^2(\textbf{X}^\prime \textbf{X})^{-1}.$$  I couldn't figure out why it is like this. Can anyone elaborate a bit?
44374
Linearity of expectation is the property that the expected value of   the sum of random variables is equal to the sum of their individual   expected values, regardless of whether they are independent.   My understanding of random variables (both continuous and discrete) is that they assign a number to each possible outcome of a random experiment.   For example, if we roll a die, we can land on any number between 1 and 6, and we can create a random variable $X$ that takes each of those values. Here, $X$ represents each possible outcome.   The expected value of $X$ would then be   $\text{E}[X] = 3.5$  by taking the weighted sum of each of the possible outcomes. It's all good until this part. Here is what I don't understand,   What is this notion of adding two random variables? I mean they don't have distinct values, so how can we say  $\text{E}[X + X] = \text{E}[X] + \text{E}[X] = 7$  This is just the linearity of expectation applied to when two dies are rolled, we are asked to get the expected sum of the numbers on both dies. But how are we adding two random variables?   What if we wanted to get the product of the number on two dies. Or the difference or quotient?   $\text{E}[X * X] = \text{E}[X] * \text{E}[X] = 12.25$  $\text{E}[X - X] = \text{E}[X] - \text{E}[X] = 0$  $\text{E}[X / X] = \text{E}[X] / \text{E}[X] = 1$   Are all of these operations valid? I'm really confused, please help. How am I supposed to think about these random variables?  This is in connection to my  that I will hopefully be able to make sense of.
98807
If $X$ is a discrete random variable with expected value $\mu$ and standard deviation $\sigma$, and $Y = X^2$, how can we describe the expected value and standard deviation of $Y$? Would expected value of $Y = \mu^2$?
7965
This should be a fairly straight forward but I still couldn't quite get it. Let $X$ be $n$ by $1$ random vector, $Y$ be another $n$ by $1$ and that $X$ and $Y$ are independent. What is $Var(X^tY)=:V_{XY}$?  I tried generalizing this formula from the univariate case $Var(xy)=E(x)^2Var(y)+E(y)^2Var(x)+Var(x)Var(y)$  except I hit a bit of a roadblock for the 3rd term above. That is, I get $$ V_{XY}=X^tV_YX+Y^tV_XY\\+E[(X^t-\mu_X^t)(Y-\mu_Y)(Y^t-\mu_Y^t)(X-\mu_X)] $$ So... what is it?
96308
I have a distribution $X$. By playing around with random samples from $X$, I've determined that $Var(X^i) &gt; Var(X)$ where $i &gt; 1$. However, I can't seem to find a formula for the expected variance of $X^i$, or why it should be greater than $Var(X)$.  Moving away from normal distributions, should it generalize that the scale parameter of any $X^i$ will be greater than $X$?
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.
56871
Can we say   $$\text{Var}(\beta_1) = \text{Var}\left(\frac{\sum (x_i-\bar x)y_i}{\sum (x_i- \bar x)^2}\right) = \left(\frac{\sum (x_i-\bar x)}{\sum (x_i- \bar x)^2}\right)^2 \text{Var}(y_i) \;\;??$$  I am not sure if I can separate the $x$'s i from $\sum (x_i-\bar x)y_i$. This expression seems like a linear combination of $y_i$'s. Is this legit because every $y_i$ follows the same distribution?
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?
24322
If X and Y are independent random variables both with the same mean (0) and variance, how about $X^2$ and $Y^2$? I tried calculating E($X^2Y^2$)-E($X^2$)E($Y^2$) but haven't been able to get anywhere.
121974
What is the mean and variance of Squared Gaussian: $Y=X^2$ where: $X\sim\mathcal{N}(0,\sigma^2)$?  It is interesting to note that Gaussian R.V here is zero-mean and non-central Chi-square Distribution doesn't work.  Thanks.
45735
According to the Page 87 in Kruschke: Doing Bayesian Data Analysis, the author says that the mean of a distribution is a value that minimizes the variance of a probability distribution, for example, a normal distribution. The following is what is mentioned in the page:  It turns out that the value of $ M $ that minimizes $ ∫dx p(x)(x−M)^2$ is $E[X] $. In other words, the mean of the distribution is the value that minimizes the expected squared deviation. In this way, the mean is a central tendency of the distribution.  I have read the paragraph and kind of understood what the author is trying to say but I wonder how this can be written mathematically using the above equation or any other ways.
72215
...assuming that I'm able to augment their knowledge about variance in an intuitive fashion (  ) or by saying: It's the average distance of the data values from the 'mean' - and since variance is in square units, we take the square root to keep the units same and that is called standard deviation.  Let's assume this much is articulated and (hopefully) understood by the 'receiver'. Now what is covariance and how would one explain it in simple English without the use of any mathematical terms/formulae? (I.e., intuitive explanation. ;)  Please note: I do know the formulae and the math behind the concept. I want to be able to 'explain' the same in an easy to understand fashion, without including the math; i.e., what does 'covariance' even mean?