To be more specific: Let the random variable X be distributed according to probability density function f(x). Then the pdf of Y=1/X is given by g(y)=1/y^2 f(1/y), .  I wonder if there is any citable reference for that. Or is it common knowledge? I'm coming from a different field of mathematics and lack some background in statistics. I looked through some text books, but didn't find any mention of it yet.  Thanks a lot!
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Suppose $X$ is a random variable with pdf $f_X(x)$. Then the random variable $Y=X^2$ has the pdf   $$f_Y(y)=\begin{cases}\frac{1}{2\sqrt{y}}\left(f_X(\sqrt{y})+f_X(-\sqrt{y})\right) &amp; y \ge 0 \\ 0 &amp; y \lt 0\end{cases}$$   I understand the calculus behind this. But I'm trying to think of a way to explain it to someone who doesn't know calculus. In particular, I'm trying to explain why the factor $\frac{1}{\sqrt{y}}$ appears out front. I'll take a stab at it:  Suppose $X$ has a Gaussian distribution. Almost all the weight of its pdf is between the values, say, $-3$ and $3.$ But that maps to 0 to 9 for $Y$. So, the heavy weight in the pdf for $X$ has been extended across a wider range of values in the transformation to $Y$. Thus, for $f_Y(y)$ to be a true pdf the extra heavy weight must be downweighted by the multiplicative factor $\frac{1}{\sqrt{y}}$  How does that sound?  If anyone can provide a better explanation of their own or link to one in a document or textbook I'd greatly appreciate it. I find this variable transformation example in several intro mathematical probability/stats books. But I never find an intuitive explanation with it :(
122555
The random variables $X_1,X_2,...X_n,Y_1,Y_2,...Y_n $ are independent and $U(0,a)$-distributed. Determine the distribution of $$Z_n=n\cdot \log\left(\frac{\max\left(X_{(n)},Y_{(n)}\right)}{\min\left(X_{(n)},Y_{(n)}\right)}\right)$$ where $X_{(n)},Y_{(n)}$ denotes $n$th order variable.  This problem is giving me a headache. The answer is that $Z_n$ is $\text{Exp}(1)$-distributed. But i can't derive that conclusion. I'd like to show you my derivation, but its so long and contains so many integrals and fractions that it would take a while to write it here. But basically my approach was to derive the distribution of $M=\max(X_{(n)},Y_{(n)})$ and $L=\min(X_{(n)},Y_{(n)})$. I get the following cdf's: $F_M=(F_X\cdot F_Y)^n=(F_X)^{2n}$ and $F_L=1-(1-(F_X)^n)\cdot (1-(F_Y)^n)=1-(1-(F_X)^n)^2$ ,  correct? $F_X $ stands for the cdf of $X_i$  Then i define the random variable $T=\frac{M}{L}$ and compute its cdf to finally be able to derive the distribution of $Z_n=n\cdot \log(T)$. But thats where i think everything goes wrong. Any help appreciated, thanks.
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I have two independent variables: $X$ follows from standard Gaussian distribution $N(0,\sigma^2)$; $Y$ follows from Rademacher distribution, i.e., $Y$ can be either $-1$ or $1$ with the same probability.  My question is:  For a new random variable of $Z=XY$, What is the probability density function?   Or can we have some concentration inequalities for $Z$ (e.g., show how $Z$ approaches its expectation, etc.)?
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Random vector $(X,Y)\sim N(0,0,1,1,\rho)$, that is to say, the density function of $(X,Y)$ is given by   $$f(x,y)=\frac{1}{2\pi\sqrt{1-\rho^2}}\exp\left\{-\frac{1}{2(1-\rho^2)}(x^2-2\rho xy+y^2)\right\}$$   Prove that    $$E\max\{X,Y\}=\sqrt{\frac{1-\rho}{\pi}}.$$   I have trouble evaluating the integral: $$\iint_{x\ge y}x\exp\left\{-\frac{1}{2(1-\rho^2)}(x^2-2\rho xy+y^2)\right\}dxdy$$ This can be turned into  $$\int_{-\infty}^{\infty}\exp\left\{-\frac{y^2}{2}\right\}dy\int_{y}^{\infty}x\exp\left\{-\frac{(x-\rho y)^2}{2(1-\rho^2)}\right\}dx$$ However, when changing variable in the second part, I cannot get rid of the $\rho y \exp\{\cdots\}$and thus got stuck here.  Any hint or solutions are welcomed, thanks!
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Let X have a logistic distribution with PDF $f_X(x)=\frac{e^{-x}}{(1+e^{-x})^2}$, $-\infty&lt;x&lt;\infty$. Show $Y=\frac{1}{1+e^{-X}}$ has an uniform distribution over (0,1).  Let $Y=u(x)=\frac{1}{1+e^{-X}} $Since this is monotone decreasing, Can use $f_Y(y)=\frac{1}{|u'(x)|}f_X(x)$.  What's the general technique of approaching?
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Question: Let $X$ and $Y$ be two independent and identically distributed exponential random variables with parameter $\lambda&gt;0$. Compute the probability density function of $X+Y$.   My Answer: I have found the joint probability density function of $X$ and $Y$ to be $f_{X,Y}(x,y)=\lambda^2e^{-\lambda x - \lambda y}$. I then let $Z=X+Y$ and calculated $F_Z(z)=\lambda^2e^{-\lambda z}$.  I know I need to integrate $F_Z(z)$ to calculate the density function but am unsure on what the limits should be or if I found $F_Z(z)$ correctly. I was also wondering if there is a quicker way to tackle such questions or if this method is okay.
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I am calculating the sum of two uniform random variables $X$ and $Y$, so that the sum is $X+Y = Z$. Since the two are independent, their densities are $f_X(x)=f_Y(x)=1$ if $0\leq x\leq1$ and $0$ otherwise. The density of the sum becomes $f_Z(z)=\int_{-\infty}^\infty f_X(z-y)f_Y(y)dy=\int_0^1f_X(z-y)dy$ by convolution. I am stuck at this stage. How do I proceed with my integral? I think a diagram make it easy but I dont know how to proceed.
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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$X,Y$ are independent, identical distributed with $$P(X=k) = P(Y=k)=\frac{1}{2^k} \,\,\,\,\,\,\,\,\,\,\,\, (k=1,2,...,n,...)$$      Calculate the probabilities $P(\min\left\{X,Y\right\} \leq x)$ and   $P(\max\left\{X,Y\right\} \leq x)$   For the minimum I do like this:  $$\begin{split}F_M(x) &amp;= P(\min\left\{X,Y\right\} \leq x) \\ &amp;= 1-P(x&lt;\min\left\{X,Y\right\} ) \\ &amp;= 1-P(x&lt;X, x&lt;Y) \\ &amp; = 1-P(X&gt;x)\,P(Y&gt;x)\\ &amp; = 1-(1-P(X \leq x))\,(1-P(Y \leq x))\\ &amp; = 1-(1-F_X(x))\,(1-F_Y(x))\end{split}$$  Is this correct for minimum? I'm not sure how do it for $\max$? Maybe I do it too complicated because they are equal these $P(X=k)=P(Y=k)$ maybe you can do it more elegant? But I don't know how?
32806
$X$ and $Y$ are independent and exponentially ditributed with parameters $a,b$, respectively. Calculate $P(X&gt;Y)$  1)$P(X&gt;Y)=1-F_{X-Y}(0)$. Then I try to calculate density  of $X-Y$ from characteristic functions but I cannot deduce distribution of $X-Y$ from this.  2)ok I can also calculate density of $Z=X-Y$ from transformation of RV formula and then marginalize  3) Is there any simpler way to calculate $P(X&gt;Y)$ than 2)?
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I've got two random variables, $x$ and $y$, where $x=u(y)$ and $y=v(x)$. How do I find the joint density of the two variables?   The eventual purpose is to find the integral  $\int^a\int^b f(x,y) dydx$,  where $f(.)$ is said joint density function.  This question is related () but I'm still a bit confused by the answer. I am indeed effectively looking for $Prob(x \leq a, y \leq b)$ but don't understand how I can get there without knowing the PDF first.  Any advice is much appreciated. Thanks a lot in advance!
16131
A random variable $X$  follows the distribution $$f_X(x)=\begin{cases}   Cx^2 &amp; -1\le x\le 2,\\   0 &amp; \text{otherwise},   \end{cases}$$ and  $Y=X^2$ . Calculate $P(X ≥ 0)$ can some one give me any hint to solve this question.
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Let $X$ and $Y$ be independent and uniform random variables on $(0,1)$. Find the density of $X+Y$.  I know $$f_{X+Y}=\int_{-\infty}^{\infty}\,f_X(x)\,f_Y(z-x)dx$$  Now when $x\in(0,1)$, I know $f_X(x)=1$. Assuming $z\in (0,1)$, I need $z&gt;x$ in order for $f_Y(y)=1$. In that case, $$f_{X+Y}=\int_{0}^{z}\,(1)\,(1)dx=z.$$   Now when $x\in(0,1)$,  $f_Y(y)$ can also equal $1$ if $z \in (1,2)$. In that case, I need $z-x&lt;1 \implies x&gt;z-1.$ So I have: $$f_{X+Y}=\int_{z-1}^{1}\,(1)\,(1)dx=2-z.$$   Does this all look correct?
73783
Let $X$ be a random variable with a geometric distribution with parameter $p$.       Calculate $\Bbb E[1/(1+X)]$   I've looked at answers online but still do not understand how to solve this.   Specifically, I do not understand how to to get from $$\frac{p}{1-p} \sum_{x=0}^\infty (1/(1+x))((1-p)^x+1)$$  to  $-p\ln(p)  /(1-p)$.
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I was reading this  where it had a gaussian distribution model.   I mean gaussian is given by  $P(y) = \frac{e^{-\frac{1}{2}(y -\mu)^T \Sigma^{-1}(y -\mu)}}{2\pi^{n/2}|\Sigma|^{1/2}}$  But is   $\frac{e^{-(y -\mu)^T \Sigma^{-1}(y -\mu)}}{2\pi^{n}|\Sigma|}$ Gaussian as well?  The paper used their model saying the gaussian had the above form. It is valid?
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Suppose that $X$ and $Y$ are chosen randomly and independently according to the uniform distribution from $(0,1)$. Define $$ Z=\frac{\max(X,Y)}{\min(X,Y)}.$$ Compute the probability distribution function of $Z$. Can anyone give me some hints on how to proceed? I can only note that $\mathbb{P}[Z\geq 1]=1$ and $$F_Z(t)= \mathbb{P}[Z \leq t]=\mathbb{P}[X\leq Y, Y\leq tX]+\mathbb{P}[Y \leq X, x\leq tY]$$
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I know very little of Probability and Statistics, and am wishing to learn.  I see the word "distribution" used all over the place in different contexts.  For example, a discrete random variable has a "probability distribution."  I know what this is. A continuous random variable has a probability density function, then for $x\in\mathbb{R}$, the integral from $-\infty$ to $x$ of the probability density function is the cumulative distribution function evaluated at $x$.   And apparently just "distribution function" is synonymous with "cumulative distribution function," at least when talking about continuous random variables (question: are they always synonyms?).  Then there are many famous distributions.  $\Gamma$ distribution $\chi^2$ distribution, etc.  But what exactly is a $\Gamma$ distribution?  Is it the cumulative distribution function of a $\Gamma$ random variable?  Or the probability density function of a $\Gamma$ random variable?  But then a frequency distribution of a finite data set appears to be a histogram.  Long story short: in Probability and Statistics, what is the definition of the word "distribution"?    I know the definition of distribution in Mathematics (an element of the dual space of the collection of test functions equipped with the inductive limit topology), but not Probability and Statistics.