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1h
comment Why is the area under the pdf for the Von Mises distribution not one?
$\exp(\cdot)$ is positive and so the pdf couldn't possibly dip below the $x$ axis. Check your code.
1h
comment Probability confusing question
Where is the seven month stretch coming from?
3h
comment Calculate the pdf from covariance matrix
Do you know that linear transformations of jointly Gaussian random variables result in jointly Gaussian random variables? Do you know what is meant by "linearity of expectation"?
2d
comment linear convolution and circular convolution
I suggest that you ask the moderators to migrate this question to he signal processing site dsp.SE. You can contact the moderators by clicking on the flag link below your question.
2d
comment Product of two multivariate Gaussian pdfs - normalization constant
If $X,Y, ZY$ are independent multivariate Gaussian random variables, the the joint density of $(X,Y,Z)$ is the product of their densities. The expression $a^TA^{-1}a+b^TB^{-1}b+c^TC^{-1}c$ together with the fact that $\exp(x)\exp(y)\exp(z) = \exp(x+y+z)$ suggests that three multivariate Gaussian densities are being multiplied together.
Apr
25
revised Probability mean,variance and standard deviation formula confusion.
added 103 characters in body
Apr
25
answered Probability mean,variance and standard deviation formula confusion.
Apr
24
comment Probability mean,variance and standard deviation formula confusion.
Read about the law of the unconscious statistician and apply it to writing formula for $E[(X-\mu)^2]$.
Apr
24
comment Determine the probability density function of…
@JohnG. You don't need to solve or compute the value of the integral as a function of $t$ in order to takes its derivative with respect to $t$. Since $t$ occurs only in the limits, just apply the Fundamental Theorem of Calculus to get the derivative directly.
Apr
24
comment $X \sim N(0, \sigma_1^2)$, $Y \sim N(0, \sigma_2^2)$, $U = X+Y$. What are $E[X|U], E[Y|U]$?
Also asked on stats.SE where it has received two answers.
Apr
23
comment If B is a N(0,1) R.V., show $E[B^4] = 3$
Hint: For $n$ even, the integral for $E[X^n]$ can be converted to a Gamma function via a change of variables.
Apr
23
comment A proof about polynomial division
If $f(x)$ has only irreducible factors of degree $2$ or more, then $g(x)$ cannot possibly divide $f^2(x)$.
Apr
23
answered P(x<1) and E(x) of a Rayleigh Distribution
Apr
22
comment DataMatrix convolutional codes
Good. Now, what are the lengths of the registers, and where are the taps? The taps might be specified as a bunch of polynomials of small degree (perhaps less than $6$ or $7$).
Apr
22
comment What is the expected value of the highest of N independent draws from the unit uniform distribution?
I didn't say that the draws are equally spaced over the interval but that one should expect the draws to be equally spaced. Suppose that you are asked to validate the behavior of a new design of a "random number generator" that returns values that_should_ be spread uniformly in $[0,1]$. Would an output that bunches all the values near $\frac 12$ not raise suspicion that something is awry? What would be more persuasive that the generator is reasonable is a sequence of $N$ values whose histogram that looks like a uniform distribution, no?
Apr
22
comment probability question, two uniformly distributed independent events, what's the probability that a third event will occur?
There are $301$ integers in the interval $[100,400]$, not $300$ as you seem to think. Also, consider that $c=b-1$ can take on values from $-300$ to $+300$, and cannot possibly be uniformly distributed. For example, $c=-300$ corresponds to just one pair of values for $(a,b)$, viz. $(a,b) = (100,400)$ whereas $c=0$ corresponds to $301$ values for $(a,b)$, ranging from $(100,100)$ to $(400,400)$.
Apr
22
comment DataMatrix convolutional codes
You need to know what the encoder looks like to make a start at a decoding algorithm. Perhaps the encoder is specified in the documentation of the "datamatrix code ECC 000-050"
Apr
22
comment The finite-dimensional distributions of a centered Gaussian process are uniquely determined by the covariance function
@MichaelHardy I didn't say that I was giving a proof. In the absence of a definition of what is understood by jointly normal random variables, what is to be proved is either a tautology (the density is $\frac{1}{(\sqrt{(2\pi)^n|\operatorname{det}(\Sigma)}}\exp(-\frac 12 \mathbf x \Sigma^{-1}\mathbf x^T)$) or a much longer proof starting from $\sum_i a_iX_i$ is normal for all choices of $a_i$ to get to the distribution depending only on the covariances.
Apr
21
comment The finite-dimensional distributions of a centered Gaussian process are uniquely determined by the covariance function
Do you know what it means for $n$ random variables to have a jointly normal distribution? If not (or perhaps even if you do), something like the next-to-last paragraph of this answer of mine over on dsp.SE might help a little.
Apr
21
comment Negative exponential distribution to the poisson distribution
In your "answer" $f_{T_1,T_2}(t_1,t_2)=\lambda^2e^{-\lambda t_2}$, where have you accounted for the fact that $T_2$ is necessarily larger than $T_1$?