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Solution for $J_1$ But what is J1? What is the expected number of steps for a frog that has only one leaf to go? The solution is $J_1 = 2(e-1)$ and other terms $J_n$ can be expressed as a sum. Rewriting the recurrence relation as a sum The recurrence relation is not gonna solve the problem entirely (because one term in the initial conditions is not ...


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The solution is: $P(Y\in [a,b]|X\in[a,b])=\frac{\int_{[a,b]}(F_{Y|X}(b|x)-F_{Y|X}(a|x))f_X(x)dx}{\int_{[a,b]}f_X(x)dx}$ This can be seen, as: $P(Y\in[a,b]\cap X\in[a,b])= P(Y\leq b\cap X\in[a,b])-P(Y\leq a\cap X\in[a,b])= \int_{a}^b \left( \int_{-\infty}^bf_{Y|X}(y|x)dy \right)dx-\int_{a}^a \left( \int_{-\infty}^af_{Y|X}(y|x)dy \right)dx= \int_a^bF_{Y|X}(...


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In order to calculate a parametric confidence intervall you need distributional assumptions. Simply using the L2 loss does not allow you to calculate confidence intervalls the "usual way". You need normally distributed errors in order for these confidence intervals to be legitimate. In that case the L2-optimizer is also the maximum likelihood estimate. So in ...


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It's simpler than it looks. To avoid writing lots of exponentials, let's work with the cumulant generating functions. These are the logarithms of the characteristic function: that is, the cgf of any random variable $X$ is $$\psi_X(s) = \log E\left[e^{isX}\right].$$ For two variables $(X,Y)$ the cgf is $$\psi_{X,Y}(s,t) = \log E\left[e^{isX+itY}\right].$$...


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A graph $\Gamma$ is an ordered pair $(V,E)$ where $V$ (the vertices) is any set and $E\subset V\times V$ (the edges) is a collection of ordered pairs of vertices. We usually depict the vertices with point symbols at distinct locations in the plane and the edges $(v,w)$ as arrows running from the location of $v$ (the origin) to the location of $w$ (the ...


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Calculate the weighted mean. If you want to give sensor2 a 70% weight, you can calculate the respective weighted means via $\text{weighted mean} = 0.7 x_{\text{sensor_2}} + 0.3 x_{\text{sensor_1}}$ Make sure the weights add up to 1!


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