Daneel Olivaw
• Member for 5 years, 7 months
• Last seen more than a month ago
• Europe

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Here is an alternative answer to @Lucas' using the law of iterated expectations: \begin{align} E\left[\sum_{i=1}^X1_{(Y_i \leq Y_{n+1})}\right] &amp; = E\left[E\left[\sum_{i=1}^X1_{(Y_i \leq Y_{n+... View answer 1 answers 1 votes 126 views Accepted answer 4 votes According to Wikipedia: A real random variable X is smaller than a random variable Y in the "usual stochastic order" if: \forall \ r \in \mathbb{R}, \ \mathbb{P}(X&gt;r) \leq \mathbb{P}(...

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You are right that the SVM decision function $w \cdot x + b$ depends only on $w$ and $b$, however it can be shown that $w$ can be expressed as a sum of support vectors. You can consult Stanford's or ...

141 views
For (1), the concern is that the program $\max_{w,b}\frac{1}{\|w\|}$ is not convex: a convex program is one in which you minimize a convex function over a convex set. See @Luca Citi 's answer for more ...
If $x$ and $y$ are independent, $\mu_x=\mu_y=0$ and $\sigma_x=\sigma_y=\sigma$, $r$ follows a Rayleigh distribution, thus: \begin{align} &amp; E[r] = \sigma \sqrt{\frac{\pi}{2}} \\[6pt] &amp; V[r] ... View answer 1 answers 0 votes 461 views Accepted answer 0 votes Let X \sim \mathcal{N}(5,2) - assuming 2 is the standard deviation - and Y = 2X+4. We have: Mean: by linearity of expectation, we get:\mathbb{E}[Y] = \mathbb{E}[2X+4] = 2\mathbb{E}[X]+4 = ...