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You're right. There's an inherent symmetry in the problem that needs to be spontaneously broken, and if you start with a symmetric guess the symmetric update procedure necessarily reproduces the symmetric guess. This not a problem, though, as long as the procedure converges to the same asymmetric solution as long as you choose asymmetric initial values.


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Suppose we have two independent range measurements $R_0,R_1$ of the range $r$ with normal noise $\nu_0(t)=\mathcal{N(0,\sigma_0^2)}$ and $\nu_1(t)=\mathcal{N(0,\sigma_1^2)}$, the Likelihood $\mathcal L(r)$ is: $$\mathcal{L}(r)=p(R_0,R_1|r)=p(R_0|r)p(R_1|r)$$ that means: ...


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First, you will need to detect the signal. For that you have several choices. You have the radar example here, so if you transmit a pulse $p(t)$, then you can apply matched filtering (which maximizes the output SNR). To do this, you need a local template which can be $p(t)$. The output of matched filter is $$x_1=\int_0^TR_1(\tau)p(t-T+\tau)d\tau,$$ which ...


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Hint: You have a likelihood of $L(\alpha\mid y_1=0) = e^{-20\alpha}$ and you also know $\alpha\ge 0$ If $e^{-20\alpha}$, which is continuous and differentiable, does not have a turning point for positive $\alpha$, then it is monotonic and will have a maximum either when $\alpha=0$ or when $\alpha \to \infty$, and it is fairly obvious which


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What you are looking at, basically, is the empirical distribution function. This is a very classical thing. According to the law of large numbers, you will eventually end up with very good estimates for $P(X < x)$. Further, by CLT, your estimator for this probability is asymptotically normal. There is also a result on the uniformity of the convergence ...


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For the $\hat{\alpha}$ part it should be quite straight forward. For $$\hat{\beta} = \frac {\bar{X}^2} {\displaystyle \frac {1} {n} \sum_{i=1}^n (X_i - \bar{X})^2}$$ you may apply continuous mapping theorem: Note that both sample mean and sample variance are consistent estimator, and thus you can claim both the numerator and denominator converge to ...


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You've already observed that $$ q(\theta) = e^{-\theta}(\theta+1) = \Pr( X_1=0 \text{ or } X_1=1). $$ So let $$ Y = \begin{cases} 1 & \text{if }X_1=0\text{ or }X_1=1, \\ 0 & \text{if }X_1\ge2. \end{cases} $$ Then $Y$ is an unbiased estimator of $q(\theta)$. You need to find $$ \operatorname{E}(Y\mid X_1+\cdots+X_n) = \Pr(Y=1\mid X_1+\cdots+X_n). $$ ...



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