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| bio | website | |
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| location | Urbana, IL | |
| age | ||
| visits | member for | 1 year, 7 months |
| seen | 1 hour ago | |
| stats | profile views | 905 |
I am a retired professor of electrical and computer engineering with a lifetime of experience teaching probability and statistics to reluctant engineering undergraduates.
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May 6 |
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Simple homework question about normally distributed variables Yes, I am sure that the variance cannot be found from the information in the current version of your question. Note that the Wikipedia page that you refer to specifically says that the two random variables are independent which word is nowhere to be found in your question. In my answer, I did point out that if one assumes that the random variables are independent, then the variances can be added up. Your $Y \sim N(0,n)$, but normality is irrelevant to the issue of the variances adding. Variances add if the variables are uncorrelated, including as a special case, independent. |
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May 4 |
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How to compute $f(R_1,R_2)$ from $f(R_1|X_1)$ and $f(R_2|X_2)$ given $X_1, X_2$ are dependent RVs? Ummm no. $R_1 = g(\cdot, Y_1,Z_1)$ and $R_2=g(\cdot,Y_2,Z_2)$ might have the same distribution or they might not, because the distribution of $R_i$ depends on the joint distribution of $Y_i, Z_i$. The marginal distributions of $Y_1$ and $Y_2$ are the same; ditto $Z_1$ and $Z_2$; but nothing said so far says that the joint distribution of $Y_1, Z_1$ is the same as the joint distribution of $Y_2,Z_2$. For example, $Y_1, Y_2, Z_1, Z_2$ are $N(0,1)$, $Y_1,Y_2$ independent, ditto $Z_1,Z_2$ but $cov(Y_1,Z_1)=0.3, cov(Y_2,Z_2)=0.7$. |
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May 4 |
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How to compute $f(R_1,R_2)$ from $f(R_1|X_1)$ and $f(R_2|X_2)$ given $X_1, X_2$ are dependent RVs? Suppose that it had turned out that $g(x,y,z)$ really is a function of only two variables $y$ and $z$ so that $R_i$ was independent of $X_i$ and $f(R_i\mid X_i) = f(R_i)$. How would you have computed $f(R_1,R_2)$ from $f(R_1)$ and $f(R_2)$ in this simpler case? |
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May 3 |
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hazard rate and cumulative incidence function If $S(t)$ is the survival function, then your calculation of CIF $= P\{T \leq t\}$ or $P\{T < t\}$ is incorrect. Note that $$P\{T \leq t\} = 1 - P\{T > t\} = 1 - S(t),$$ so that no integration, and especially not an integration of $h(\cdot)S(\cdot)$, is needed for the calculation of the CIF from the survival function. |
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May 3 |
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hazard rate and cumulative incidence function What is $S$? $h(t)$ is a function whose integral over the positive real line diverges and what you call the CIF has a different relation to $h(t)$. In particular, $$P\{T \leq t\}=1 - \exp\left(-\int_0^t h(u)\,\mathrm du\right),$$ where the integrand is the area under $h(\cdot)$ between $0$ and $t$. As a special case, constant hazard rate $h(t)=\lambda$ makes $T$ work out to be an exponential random variable with mean $\lambda^{-1}$. |
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Apr 26 |
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Can we assume that the true variance is equal to observed variance minus 2 multiplied by (cov)2 Could you state the variance covariance theorem that you are asking about? |
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Apr 26 |
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Moment generating functions question The issue here is that the Answer given to Sue is not an answer to the question she has been asked: What is the distribution of their sum: $\sum_{i=1}^n X_i$? The correct answer to the question that she has been asked is that the sum has a Poisson distribution with parameter $\lambda=\sum_{i=1}^n \lambda_i$ and not The moment generating function of Poisson(sum of $\lambda$ ); that is the answer to a different question: What is the MGF of $\sum_{i=1}^n X_i$? |
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Apr 20 |
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Why is it called white noise? The power spectral density of a white noise process (random process) has constant value for all frequencies. Someone thought that this was a property of white light (equal energy at all frequencies (in a certain range of frequencies)) and named the random process white noise by analogy. People liked the name and it stuck. In fact, white light has equal energy at all wavelengths, not frequencies, but it is too inconvenient to correct the name now. |
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Apr 19 |
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Is there a function to check a set of events/random variables are mutually independent? I do not claim to be a statistician nor do I play one on the Internet, but if your list has the exact probabilities instead of data obtained by simulation, then at least in this instance, an eyeball test shows that $E$ is independent of the other random variables since the joint probability $P(A,B,C,D,E)$ equals $0$ whenever $E=1$. So it suffices to check independence of $A,B,C,D$ only which can be done by finding the marginal mass functions (column sums!) and verifying that $$p_{A,B,C,D}(a,b,c,d)=p_A(a)p_B(b)p_C(c)p_D(d), a,b,c,d \in \{0,1\}$$ on all $16$ nonzero rows of your table. |
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Apr 19 |
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Statistics of a simple coin toss According to LaPlace's rule of succession, the conditional probability of getting a Head on the next toss, given that the first $100$ tosses resulted in Heads is $\frac{101}{102}$. The coin does not have to fair any more than the sun is. |
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Apr 18 |
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What's the distribution of $\bar{X}^{-1}$? @cardinal (continued) the density is $0$ on $(a^{−1},b^{−1})$ for all $n$. |
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Apr 18 |
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What's the distribution of $\bar{X}^{-1}$? @guy You are correct. I am deleting my previous comment except for the first few words. |
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Apr 18 |
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What's the distribution of $\bar{X}^{-1}$? @cardinal Yes, I understand that. But the OP asked two questions: (i) what is the distribution of $\bar{X}^{-1}$? and (ii) can the CLT be used in getting the distribution? I think the answer to the second question is No. The point brought up guy in comments, not in his answer and further explicated by you says that the SLLN, not the CLT shows that $\bar{X}^{-1}$ converges a.s. to $\mu^{-1}$. Indeed, the whole delta method and CLT invites the inference (by casual readers or beginners) that for (fixed) large $n$, $\bar{X}^{-1}$ is approximately normal while we know the density is $0$ |
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Apr 16 |
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Question about sample autocovariance function If your time series is exactly $x_1, x_2, \ldots, x_n$ with all other $x_i$, $i < 1$ or $i >n$ being unknown, then the sum must necessarily stop at $t=n-h$ when $x_{t+h}=x_n$ occurs in the sum: the next term (for $t=n-h+1$) that would be included in the sum would have $x_{n-h+1+h}=x_{n+1}$ in it, and $x_{n+1}$ is not part of the sample. |
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Apr 16 |
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What's the distribution of $\bar{X}^{-1}$? I am not sure if the conditions for SLLN are satisfied. Does $\bar{X}^{-1}$ have a mean? Since the density of $\bar{X}$ is nonzero and continuous at $0$ for all $n$ (when $a < 0 < b$), I suspect that it does not. |
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Apr 16 |
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What's the distribution of $\bar{X}^{-1}$? The question was about the distribution of $\bar{X}$. For $X\sim \mathcal U(a,b)$, $\bar{X}$ takes on values in $(a,b)$ for all $n$. If $a < 0 < b$, the support of the density of $\bar{X}^{-1}$ is $(-\infty, a^{-1}) \cup (b^{-1}, \infty)$ for all $n$. So, what is the distribution of $\bar{X}^{-1}$ converging to? Specifically, for $\mathcal U(-2,1)$, is the density of $\bar{X}^{-1}$ converging to a unit probability mass at $-2$, the reciprocal of $\mu = (-2+1)/2 = -\frac{1}{2}$? |
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Apr 14 |
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What's the distribution of $\bar{X}^{-1}$? Are you sure you know exactly what the CLT says, and whether it is at all applicable to the sample mean $\bar{X}$? |
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Apr 12 |
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Expected values of (X*Y^2) when X and Y are dependent normally distributed RVs Assuming that $X$ and $Y$ are jointly normal (as I included in my edit), vinux's hint is the way to go. Given the value of $X$, $Y$ is a normal random variable whose mean and variance are known, and so $$E[Z\mid X] = E[XY^2\mid X] = XE[Y^2\mid X] = X(\sigma_{Y\mid X}^2 + \mu_{Y\mid X}^2 = g(X)$$ where $g(X)$ is a cubic in $X$ and so only the quadratic and constant term matter: $E[X] = E[X^3] = 0$ in this case. |
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Apr 11 |
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Finding Expected Value of a discrete uniform random variable >Isn't this pretty much the binomial distribution? Well, to parody a quote from an ex-President, it pretty much depends on what the meaning of pretty much is. |
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Apr 11 |
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Finding Expected Value of a discrete uniform random variable If $1 < a \leq \theta$ and exactly one or two or three or ... or $n$ of the $X_i$'s equal $a$ while the remaining $X_j$ are smaller than $a$, then it must be that $\max_i X_i = a$, right? So, maybe $$P\{\max_i X_i = a\} = \sum_{k=1}^n \binom{n}{k}\left(\frac{1}{\theta}\right)^k\left(\frac{a-1}{\theta}\right)^{n-k},~~ 1 < a \leq \theta ?$$ I will leave it to you to work out $P\{\max_i X_i = 1\}$. |
