I understand that the integral of a PDF provides tangible value --i.e., the integral of a PDF allows one to see the probability of a value or less than that value, under a particular distribution, occurring. But, what does the value of just the output of the PDF provide? In other words, what does the PDF of the standard normal distribution at x=0.5 mean?
7429
I understand that the probability mass function of a discrete random-variable X is $y=g(x)$. This means $P(X=x_0) = g(x_0)$.  Now, a probability density function of of a continuous random variable X is $y=f(x)$. Wikipedia defines this function $y$ to mean      In probability theory, a probability density function (pdf), or density of a continuous random variable, is a function that describes the relative likelihood for this random variable to take on a given value.   I am confused about the meaning of 'relative likelihood' because it certainly does not mean probability! The probability $P(X&lt;x_0)$ is given by some integral of the pdf.  So what does $f(x_0)$ indicate? It gives a real number, but isn't the relative likelihood of a specific value for a CRV always zero?
95311
If $X \sim \exp(3)$ and let $Y = e^{2X}$ what is the PDF of $Y$?  $$ \begin{align} F_Y(y)&amp; = P(Y\leq y) \\ &amp; = P(e^{2X} \leq y) \\ &amp;=P(X \leq \ln(\sqrt{y})) \end{align} $$ Then you need your CDF will look like $$ \begin{align} P(X \leq \ln(\sqrt{y})) &amp;= \int_0^{\ln(\sqrt{y})}3e^{-3x}dx \\ \end{align} $$ Using the fundamental theorem of calculus we get that the PDF of Y will be  $$ \begin{align} f_Y(y) &amp;= 3(e^{-3ln(\sqrt{y})}-1) \\ &amp; =3(y^{-3/2}-1) \quad y&gt;1  \end{align} $$  The PDF doesnt integrate to 1 on it's support so there has to be a mistake however I don't where exactly please help me find it.
45080
Suppose $\phi(\cdot)$ and $\Phi(\cdot)$ are density function and distribution function of the standard normal distribution.  How can one calculate the integral:  $$\int^{\infty}_{-\infty}\Phi\left(\frac{w-a}{b}\right)\phi(w)\,\mathrm dw$$
14755
If the Standard Normal PDF is $$f(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}$$  and the CDF is  $$F(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-x^2/2}\mathrm{d}x\,,$$  how does this turn into an error function of $z$?
114727
Given $$f(x) = \begin{cases} x, \; 0 &lt; x &lt;1\\ 2-x, \; 1 \leq x &lt; 2\\ 0 \text{ everywhere else} \end{cases}$$  as our P.D.F, I must find the corresponding distribution function.   I know that $F(x) = P(X \leq x) = \int_{-\infty}^{x}f(t) dt$ is the distribution function, but I don't know how to apply it to this particular probability density function. The P.D.F has three different cases. How do I handle that? A sum of integrals with the appropriate bounds for each case of $f(x)$?  I'm kind of confused on how to create those bounds. How do we get rid the of the $-\infty$?  I'm thinking:  For $x &gt; 0$:  $$F(x) = P(X \leq x) = \int_{-\infty}^{x} f(t)dt = \int_{0}^{1} x dx + \int_{1}^{2} 2-x dx$$ Eh....
76134
I got a problem of calculating $E[e^X]$, where X follows a normal distribution $N(\mu, \sigma^2)$ of mean $\mu$ and standard deviation $\sigma$.  I still got no clue how to solve it. Assume $Y=e^X$. Trying to calculate this value directly by substitution $f(x) = \frac{1}{\sqrt{2\pi\sigma^2}}\, e^{\frac{-(x-\mu)^2}{2\sigma^2}}$ then find $g(y)$ of $Y$ is a nightmare (and I don't know how to calculate this integral to be honest).  Another way is to find the inverse function. Assume $Y=\phi(X)$, if $\phi$ is differentiable, monotonic, and have inverse function $X=\psi(Y)$ then $g(y)$ (PDF of random variable $Y$) is as follows: $g(y)=f[\psi(y)]|\psi'(y)|$.  I think we don't need to find PDF of $Y$ explicitly to find $E[Y]$. This seems to be a classic problem. Anyone can help?
110892
Since integration is not my strong suit I need some feedback on this, please:  Let $Y$ be $\mathcal{N}(\mu,\sigma^2)$, the normal distrubution with parameters $\mu$ and $\sigma^2$. I know $\mu$ is the expectation value and $\sigma$ is the variance of $Y$.  I want to calculate the $n$-th central moments of $Y$.  The density function of $Y$ is $$f(x)=\frac{1}{\sigma\sqrt {2\pi}}e^{-\frac{1}{2}\left(\frac{y-\mu}{\sigma}\right)^2}$$  The $n$-th central moment of $Y$ is $$E[(Y-E(Y))^n]$$  The $n$-th moment of $Y$ is $$E(Y^n)=\psi^{(n)}(0)$$ where $\psi$ is the Moment-generating function $$\psi(t)=E(e^{tX})$$  So I started calculating:  $$\begin{align}  E[(Y-E(Y))^n]&amp;=\int_\mathbb{R}\left(f(x)-\int_\mathbb{R}f(x)dx\right)^n\,dx \\ &amp;=\int_\mathbb{R}\sum_{k=0}^n\left[\binom{n}{k}(f(x))^k\left(-\int_\mathbb{R}f(x)dx\right)^{n-k}\right]\,dx \\ &amp;=\sum_{k=0}^n\binom{n}{k}\left(\int_\mathbb{R}\left[(f(x))^k\left(-\int_\mathbb{R}f(x)dx\right)^{n-k}\right]\,dx\right) \\ &amp;=\sum_{k=0}^n\binom{n}{k}\left(\int_\mathbb{R}\left[(f(x))^k\left(-\mu\right)^{n-k}\right]\,dx\right) \\ &amp;=\sum_{k=0}^n\binom{n}{k}\left((-\mu)^{n-k}\int_\mathbb{R}(f(x))^k\,dx\right) \\ &amp;=\sum_{k=0}^n\binom{n}{k}\left((-\mu)^{n-k}E\left(Y^k\right)\right) \\ \end{align}$$  Am I on the right track or completely misguided? If I have made no mistakes so far, I would be glad to get some inspiration because I am stuck here. Thanks!
84706
I'm solving a problem where I've this 'expectation': $$ \int_{0}^y x\cdot f(x) dx $$ where $f(x)$ is a PDF with support on $[0, z]$, with $z&gt;y$. Is there a way to rewrite it without the integral and as a function of the CDF? I've tried integration by parts, but without great success: $$ \int_{0}^y x\cdot f(x) dx = y\cdot F(y) -\int_0^y F(x) dx $$ I have hard time to solve the second part.
29256
Inspired by my other , I would like to ask how does one find the mode of a probability density function (PDF) of a function $f(x)$?  Is there any "cook-book" procedure for this? Apparently, this task is much more difficult than it seems at first.
15959
Equivalently, about variance?  I realize it measures the spread of a distribution, but many other metrics could do the same (e.g., the average absolute deviation). What is its deeper significance? Does it have    a particular geometric interpretation (in the sense, e.g., that the mean is the balancing point of a distribution)?  any other intuitive interpretation that differentiates it from other possible measures of spread?   What's so special about it that makes it act as a normalizing factor in all sorts of situations (for example, convert covariance to correlation)?
62388
How can we derive the chi-squared probability density function (pdf) using the pdf of normal distribution?  I mean, I need to show that   $$f(x)=\frac{1}{2^{r/2}\Gamma(r/2)}x^{r/2-1}e^{-x/2} \&gt;, \qquad x &gt; 0\&gt;.$$
29376
I mean statements such as: "the probability distribution of X" or "its probability distribution". I am quite confused about such expressions.  Could someone please explain the relationship between, Probability, Probability Distribution, density function? I have checked on Wikipedia for a while, still only clear about that, density function can be used to calculate the probability.
75298
To understand what I mean, let's use two well-known distributions: the normal and lognormal ones.  From the dataset point of view, if you take normally-distributed data and take their exponential, you get lognormally-distributed data (norm -exp-> lognorm).  So the "idea" was: why not using the PDF directy and transform the normal PDF to the lognormal one by taking the exponential of its x-axis. The reasoning was that this x-axis represents my data: taking the exponential of these data or the exponential of the x-axis of the corresponding PDF should do the same.  But I did not even bother to do it: mathematically, this is not how one can transform the normal PDF to the lognormal one!  What I don't understand is why just taking the exponential of the x-axis of the PDF does not work. I mean, not mathematically (I get that there is a $1/x$ in the lognormal PDF!), but what is conceptually wrong with "just stretching" the axis?
103675
I've been going back over my notes from Stats class and came across the Probability Integral Transform.  From my limited understanding, the basic idea is that from a cdf in terms of one variable, can be transformed into another cdf in terms of different variable:    i.e. from $F_x(x)$ to --> $F_y(y)$   Is this understanding correct?  What is the purpose behind this?  Finally, is there a general procedure in performing the transformation?
49358
It's a little tricky to me, I am not sure how to proceed...my atempt was to stick strictly to the definitions given and the formula for the (continuous) expected value.  Prove that $E[e^{\mu+\sigma Z}\mathbb{1}_{Z&gt;-d}]=e^{\mu+\frac{\sigma ^2}{2}}\Phi(d+\sigma)$, where $\Phi$ is the standard normal distribution function, $Z  \sim N(0,1)$(i.e. standard normal distribution) and $\mathbb{1}_{Z&gt;-d}$ denotes the following indicator function of an event such that, $$\mathbb{1}_{Z&gt;-d}=1 \text{ when } Z &gt;-d \text{ and, } \mathbb{1}_{Z&gt;-d}=0 \text{  otherwise}$$ during your work, denote the standard normal density function as $\phi$.  I only got up to thinking $e^{\mu+\sigma Z}\mathbb{1}_{Z&gt;-d}$ is essentially just $e^{\mu+\sigma Z}$ if $Z&gt;-d$ for some $d$ and $0$ otherwise. And so, the expected value is defined (for continuous random variables) $E[X]= \int_{-\infty}^{\infty}xf(x)dx$ where $f(x)$ is the probability for each random variable $x$. I thought this would lead me to using $\phi$ and ultimately leading me to $\Phi$ in my final conclusion as in the RHS. But I am not sure about how to proceed, would someone please help me? It would be great if you could explain the steps to me...thank you very much in advance...
96676
Suppose $X$ and $Y$ are independent random variables on $\mathbb{R}$ having pdf $f(t)=\frac{1}{\pi} \frac{1}{1+t^2}$. Define $Z=\frac{X+Y}{3}$ determine the pdf of $Z$.   So the pdf of $(X,Y)$ is $f(x,y)=\frac{1}{\pi^2 (1+x^2)(1+y^2)}$.   We shall find the CDF of $Z$ and then differentiate it.  $F(z)=P(Z \leq z) = \int_{-\infty}^{\infty} \int_{-\infty}^{3z-x} \ \frac{1}{\pi^2 (1+x^2)(1+y^2)} \  dy \ dx= \int_{-\infty}^{\infty} \frac{1}{\pi^2}\left(\frac{\pi}{2(1+x^2)}- \frac{\tan^{-1}(x-3z)}{1+x^2}\right) \ dx$.  At this stage I am having trouble evaluating $\int_{-\infty}^{\infty} \left(\frac{\tan^{-1}(x-3z)}{1+x^2}\right) \ dx$.  How do I do this? Or otherwise is there different more elegant way to get the required pdf?
33948
The density function for a random variable X is given in terms of a constant $c$.  Find the value of $c$.  What is the corresponding distribution function?  Sketch both the density and the distribution functions. Finally, find the probabilities.  5.1 $f(x)=0$ for $x&lt;0$ and $f(x)=\frac{c}{(x+1)^4}$ for $0&gt;x$   $P(X&gt;4)$  $P(X&lt;2)$  $P(1\le\ X&lt;3)$   I get $c=3$ by taking the integral, and then I am a little confused as to why  $F(x)=1-\frac{1}{(x+1)^3}$  Is that because in solving for $c$, I ended up with $1=\frac{c}{3(x+1)^3}$, so substituting c in I get $1=\frac{1}{(x+1)^3}$?  And, then  $P(X&gt;4)=1-P(X\le\ 4)$ $=1-(1-{1}{(1+4)^3})$ $=1/125$  $P(X&lt;2)=1-{1}{(1+2)^3}$ $=26/27$  I guess I am a little confused about what $F(x)$ and $f(x)$ actually MEAN? And thus that leads to my confusion about what the differences between $&gt;$ and $&lt;$ and greater than or equal to are, etc etc in terms of the equations. The ones above I solved by comparing to the book but I still don't understand what they mean really.  So as a result, I don't really get how to solve $P(1\le\ X&lt;3)$.  5.2 $f(x)=ce^x$ for $x&lt;0$ and $f(x)=ce^{-x}$ for $0&gt;x$  5.3 $f(x)=0$ for $x&lt;-1$ or $x&gt;2$ and $f(x)=cx^2$ for $-1&lt;x&lt;2$