Say, for example, we have a $Y \sim \mathrm{Uniform}(0, 0.25)$, then $P(Y = y) = 4$ for all $Y \in [0, 0.25]$. My question is: what is the statistical significance/meaning of the $4$?
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On the , there is this line:      $p(\mathrm{height}|\mathrm{male}) = 1.5789$ (A probability distribution over 1 is OK. It is the area under the bell curve that is equal to 1.)    How can a value $&gt;1$ be OK? I thought all probability values were expressed in the range $0 \leq p \leq 1$. Furthermore, given that it is possible to have such a value, how is that value obtained in the example shown on the page?
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Say I have $X \sim \mathcal N(a, b)$ and $Y\sim \mathcal N(c, d)$.  Is $XY$ also normally distributed?  Is the answer any different if we know that $X$ and $Y$ are independent?
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Given three Uniform random variables between 0 and 1, $(x,y,z)$, that are i.i.d., what is the probability $x+y+z &lt; 1$?
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In my statistics class, we're just beginning to talk about (point) estimation. I have a small question that might actually be due more to notation/definition than anything very conceptual:  Say you have a population $P$ of students in a classroom and you want to find the mean height by randomly selecting $n$ students $s_1,...,s_n$ in $S\subseteq P$ to measure. My book calls $X_1,...,X_n$ a random sample. I'm just a little confused as to what $X_i$ is defined to be (what $X_i$, as a random variable, is mapping from). In this example, my book would say {$X_1,...,X_n$} are the respective heights of the $n$ randomly selected students. Just to make sure I have the right picture in my head: $S$ is the ("first") sample space and $X:S\to\mathbb{R}$ is a random variable ($X$ is the "height" function defined on $S$). We denote $X_i$ as the restriction of $X$ to the $s_i$-th student in $S$. That is, we define $X_i:${$s_i$}$\to\mathbb{R}$ so $X_i(s_j)$ is not defined for $i\neq j$, and:   $X_1(s_1)=X(s_1)=x_1$  $X_2(s_2)=X(s_2)=x_2$  $.$  $.$  $.$  $X_n(s_n)=X(s_n)=x_n,$ where $x_1,...,x_n$ are the respective heights of the $n$ students. Then $Range(X)=${$X_1(s_1),...,X_n(s_n)$}$=${$X_1,....,X_n$} is considered the new sample space that we define the pmf/pdf $f$ on, where $f:\mathbb{R}\to\mathbb{R}$ and $f\circ X(s_1)=f(x_i)$.   Is this the right idea?
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$X,Y$ and $Z$ are independent uniformly distributed on $[0,1]$  How is random variable $(XY)^Z$ distributed?  I had an idea to logarithm this and use convolution integral for the sum, but I'm not sure it's possible.
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I was implementing a simple scheme of Bernoulli distribution sampler. $ X \sim B(p) $. I have a function that generates a uniform random number $r \in (0,1)$. Then, I set $ X = 1 $  if $p &gt; r $, and $X =0$ otherwise. Is this correct?
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Let $X$ be a random variable that follows $Uniform(-1,2)$ distribution.   Then what will be the cdf of $X^2$?   What i did was $$P(X^2\le y)=P(X\le y^{0.5})= \int_{-1}^{y^{0.5}}(1/3)dx=(1/3)(y^{0.5}+1) $$ $0\lt y \lt 4$ But this cdf doesnt give the probability $0$ at $y=0$ also the pdf which i get doesn't integrate to 1 What is wrong can somebody explain?? Am i doing something wrong??
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According to Bayes' theorem, $P(y|\theta)P(\theta) = P(\theta|y)P(y)$. But according to my econometric text, it says that $P(\theta|y) \propto P(y|\theta)P(\theta)$. Why is it like this? I don't get why $P(y)$ is ignored.
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I have been reading Geoff Cumming's 2008 paper  [~200 citations in Google Scholar] -- and  am confused by one of its central claims. This is one in the series of papers where Cumming argues against $p$-values and in favour of confidence intervals; my question, however, is not about this debate and only concerns one specific claim about $p$-values.   Let me quote from the abstract:     This article shows that, if an initial   experiment results in two-tailed $p= .05$, there is an $80\%$   chance the one-tailed $p$-value from a replication will fall in   the interval $(.00008, .44)$, a $10\%$ chance that $p &lt; .00008$,   and fully a $10\%$ chance that $p &gt; .44$. Remarkably, the   interval—termed a $p$ interval—is this wide however large   the sample size.   Cumming claims that this "$p$ interval", and in fact the whole distribution of $p$-values that one would obtain when replicating the original experiment (with the same fixed sample size), depend only on the original $p$-value $p_\mathrm{obt}$ and do not depend on the true effect size, power, sample size, or anything else:     [...] the probability distribution of $p$ can be derived without   knowing or assuming a value for $\delta$ (or power). [...] We do not assume any prior knowledge   about $\delta$, and we use only the information $M_\mathrm{diff}$ [observed between-group difference] gives about $\delta$ as   the basis for the calculation for a given $p_\mathrm{obt}$ of the distribution of   $p$ and of $p$ intervals.      $\quad\quad\quad$    I am confused by this because to me it seems that the distribution of $p$-values strongly depends on power, whereas the original $p_\mathrm{obt}$ on its own does not give any information about it. It might be that the true effect size is $\delta=0$ and then the distribution is uniform; or maybe the true effect size is huge and then we should expect mostly very small $p$-values. Of course one can start with assuming some prior over possible effect sizes and integrate over it, but Cumming seems to claim that this is not what he is doing.  Question: What exactly is going on here?    Note that this topic is related to this question:  with an excellent answer by @whuber. Cumming has a paper on this topic to:  -- but that one is clear and unproblematic.  I also note that Cumming's claim is repeated several times in the 2015 Nature Methods paper  that some of you might have come across (it already has ~100 citations in Google Scholar):     [...] there will be substantial variation in the $P$ value of repeated experiments. In reality, experiments are rarely repeated; we do not know how different the next $P$ might be. But it is likely that it could be very different. For example, regardless of the statistical power of an experiment, if a single replicate returns a $P$ value of $0.05$, there is an $80\%$ chance that a repeat experiment would return a $P$ value between $0$ and $0.44$ (and a $20\%$ change [sic] that $P$ would be even larger).   (Note, by the way, how, irrespective of whether Cumming's statement is correct or not, Nature Methods paper quotes it inaccurately: according to Cumming, it's only $10\%$ probability above $0.44$. And yes, the paper does say "20% change". Pfff.)
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Given two independent random variables $X \sim \mathcal{U}[-1,5]$ and $Y \sim \mathcal{U}[-5,5]$, what is $$\mathbb{E}\{Y - X | X \le 1, Y &gt; X, Y \in [-1,1] \}\,?$$  I managed to compute the respective probability   $$ \mathbb{P}\{ X \le 1, Y &gt; X, Y \in [-1,1] \} = \mathbb{P}(X \le 1) \mathbb{P}(Y &gt; X) \mathbb{P}(Y \in [-1,1]) = 1/3 * 1/2 * 1/5 = 1/30$$  How do I continue from there?
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Suppose that we have the situation as depicted in the figure: a random experiment which has 4 outcomes $x_1, ..., x_4$ and two events $A$ and $B$. Also assume that $P(x_i)=0.25$.  Now, since $P(A \cap B) = P(A)P(B) = 0.25$, by definition the events $A$ and $B$ are independent. This does not make sense to me. Why are they independent? Is there an intuitive explanation?  Another thing is, suppose that there is a 5th outcome, $x_5$, outside of $A \cup B$, in that case the events $A$ and $B$ are no longer independent since  $P(A \cap B) \ne P(A)P(B)$. This result also does not make sense to me.
43115
Suppose that  $X$ and $Y$ are two i.i.d. uniform random variables on the interval $[0,1]$  Let $Z=X/Y$, I am finding the cdf of $Z$, i.e. $ \Pr(Z\leq z) $.  Now, I came up with two ways of doing this. One produces a correct answer consistent with the pdf here: , the other does not. Why is the second method wrong?  First Method  $\newcommand{\rd}{\mathrm{d}} \Pr(Z\leq z) = \Pr(X/Y\leq z) = \Pr(X\leq zY) = \int^{1}_{0}\int^{\min(1,zy)}_{0} \rd x \rd y = \int^{1}_{0}\min(1,zy)\ \rd y$ $  = \left\{      \begin{array}{lr}        \int^{1/z}_{0}zy\ \rd y + \int^{1}_{1/z} \rd y&amp; : z &gt; 1\\        \int^{1}_{0}zy\ \rd y &amp; : z \leq 1      \end{array}    \right. $ $  = \left\{      \begin{array}{lr}        1 - \frac{1}{2z} &amp; : z &gt; 1\\        \frac{z}{2} &amp; : z \leq 1      \end{array}    \right. $  This appears correct.  Second Method  $\Pr(X/Y\leq z) = \Pr(X \leq zY\ |\ zY \geq 1)\Pr(zY \geq 1) + \Pr(X \leq zY\ |\ zY &lt; 1)\Pr(zY &lt; 1) $ by total probability  $ = \Pr(X \leq zY\ |\ zY \geq 1)\Pr(Y \geq 1/z) + \Pr(X \leq zY\ |\ zY &lt; 1)\Pr(Y &lt; 1/z)$  Taking $z&gt;1$ yields $ (1)(1-\frac{1}{z}) + (\int^{1/z}_{0}\int^{zy}_{0} \rd x \rd y)(\frac{1}{z}) = 1-\frac{1}{z} + (\int^{1/z}_{0}zy\ \rd y)(\frac{1}{z}) = 1-\frac{1}{z} + \frac{1}{2z^{2}}$  This is already different. Why is this wrong?  Thanks!
9295
I'm trying to follow a line in a derivation for $P(Z&gt;X+Y)$ where $X,Y,Z$ are independent continuous random variables distributed uniformly on $(0,1)$.  I've already derived the pdf of $X+Y$ using the convolution theorem, but there's a line in the answer that says:  $P(Z&gt;X+Y) = \mathbb{E}[\ P(Z&gt;X+Y\ |\  X+Y )\  ]$ where $\mathbb{E}$ is the expectation.  I'm not familiar with this result. Could anyone give a pointer to a similar result if one exists?  Thanks.
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Help me to solve this problem please..  Let $Y_{(1)}, Y_{(2)}, Y_{(3)}, Y_{(4)}, Y_{(5)}$ denote the order statistics of a random sample of size 5 from a distribution having p.d.f. $f(y) = e^{(-y)}, 0 &lt; y &lt; \infty$, zero elsewhere. Show that $Z_1 = Y_{(2)}$ and $Z_2 = Y_{(4)} − Y_{(2)}$  are independent. Hint: First find the joint p.d.f. of $Y_{(2)}$  and  $Y_{(4)}$.
12085
The following is a homework question for which I am asking guidance.     Let $A$, $B$, $C$ be independent random variables uniformly distributed between $(0,1)$. What is the probability that the polynomial $Ax^2 + Bx + C$ has real roots?   That means I need $P(B^2 -4AC \geq 0$). I've tried calling $X=B^2 -4AC$ and finding $1-F_X(0)$, where $F$ is the cumulative distribution function.  I have two problems with this approach. First, I'm having trouble determining the product of two uniform random variables. We haven't been taught anything like this in class, and couldn't find anything like it on Sheldon Ross' Introduction to Probability Models.  Second, this strategy just seems wrong, because it involves so many steps and subjects we haven't seen in class. Even if I calculate the product of $A$ and $C$, I'll still have to square $B$, multiply $AC$ by four and then subtract those results. It's too much for a homework question. I'm hoping there might be an easier way.
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I know how to use PDFs to calculate probabilities, but I don't think I understand them. For example, at $X=0$ the PDF of the standard normal distribution is $\approx 0.4$. Does this have any useful meaning?
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Let $X$ and $Y$ be independent random variables, uniformly distributed in the interval $[0,1]$. Find the CDF and the PDF of $|X - Y|$?  Attempt Let $Z = |X - Y|$, so for $z \geq 0$, the CDF $F_{Z}(z) = \mathbf{P}(Z \leq z) = \mathbf{P}(|X - Y| \leq z) = \mathbf{P}(-z \leq X - Y \leq z)$, which is where the algebra becomes confusing. Since they are independent, the joint pdf of $X$ &amp; $Y$ is simply 1, as long as $(X,Y)$ belong to the unit square. The solution suggests a plot the event of interest as a subset of the unit square and find its area. Any hints?