If $X = N_x$ and $Y = 4*X + N_y$, where $N_x$, $N_y \sim N(0, 1)$, then we have that $P(Y|X=x) = N(4x, 1)$. If we want to then find $P(X|Y=y)$, I would think that Bayes would work:  $P(X|Y=y) = P(Y|X=x)P(x)/P(y) = N(4x, 1)P(x)/P(y)$.   For x=2, the solution is N(8/17, 1/17). I don't see how this results. Can someone clarify please?  EDIT: Another user cited this as being a duplicate, but I don't see why.   EDIT: Originally, I asked "what is $P(x)$, $P(y)$? Are they the pdf of $N(0, 1)$ at their respective values?"
118606
How do I complete the square from the point I have left off at, and is this correct so far?   I have a normal prior for $\beta$ of the form $p(\beta|\sigma^2)\sim \mathcal{N}(0,\sigma^2V)$, to get:  $p(\beta|\sigma^2)=(2\pi\sigma^2V)^\frac{p}{2}\exp[-\frac{1}{2\sigma^2}\beta^T\beta]$  where $\beta^T\beta$ is   $\sum\limits_{i=1}^p \beta_i^2$.  My likelihood has a normal distribtuion for the data points y of the form $p(y|\beta,\sigma^2)\sim\mathcal{N}(B\beta,\sigma^2I)  $  $p(y|\beta,\sigma^2)=(2\pi \sigma^2V)^\frac{n}{2}\exp[-\frac{1}{2\sigma^2}({\bf y}-{\bf B}{\bf \beta})^T({\bf y}-{\bf B}{\bf \beta})]$  (Note that $\beta$ is a matrix/vector too, \bf does not work.)  To get my posterior for $\beta$ I combined the above, took the exponential parts only, and then expanded to get:  $\exp[-\frac{1}{2\sigma^2}({\bf y}^T{\bf y}-{\bf y}^T{\bf B}\beta-\beta{\bf B}^T{\bf y}-\beta^T{\bf B}^T{\bf B}\beta)]\exp[-\frac{1}{2\sigma^2}({\bf \beta}^T{\bf B})]$.  I dropped the $({\bf y}^T{\bf y})$ term, as is not a function of $\beta$.  Putting into one expression without the exponential:  $-\frac{1}{2\sigma^2}(-{\bf y}^T{\bf B}\beta-\beta{\bf B}^T{\bf y}-\beta^T{\bf B}^T{\bf B}\beta+{\bf \beta}^T{\bf B})$.  I know I need to combine the similar terms and get into the form of the multivariate normal distribution, which is what I am aiming for, but I am unsure how to do this? I probably have to add an extra term to the expression to get it into the correct form?  Note: This is not homework, it's a project, but my Bayesian working knowledge is not good at all and so I need to understand the working out. I intend to integrate out the $\beta$ and then the $\sigma^2$ after getting into the multivariate form.
120996
If $X$ and $Y$ are independent geometric random variables, $X \sim G(p)$ and $Y \sim G(q)$ then if $Z = X+Y$ I need to show that:     $$P[Z=z] = \frac{pq}{p-q}[(1-q)^{z-1}-(1-p)^{z-1}]$$   My attempt: \begin{align} P[Z=z] &amp;= \sum_{k=1}^z P[X=k]P[Y=z-k] &amp;&amp; \text{(not sure of my summation limits here)}  \\ &amp;= \sum_{k=1}^z p(1-p)^{k-1} \cdot q(1-q)^{z-k-1} \\ &amp;= \frac{pq}{(1-p)}\cdot \frac{(1-q)^z}{(1-q)}\cdot\sum_{k=1}^z (\frac{1-p}{1-q})^k \\ &amp;= \frac{pq}{(1-p)}\cdot \frac{(1-q)^z}{(1-q)} \cdot \frac{1-(\frac{1-p}{1-q})^k}{1-\frac{1-p}{1-q}} \\ &amp;= \frac{pq}{(1-p)}\cdot \frac{(1-q)^z}{(p-q)} \left[1 -(\frac{1-p}{1-q})^k\right] \\ &amp;= \frac{pq}{p-q}\left[\frac{(1-q)^z}{1-p} - \frac{(1-p)^z}{1-p}\right] \end{align}  The result is pretty close to the answer. The only discrepancy is the first expression in the parentheses, $\frac{(1-q)^z}{1-p}$ which should be $\frac{(1-q)^z}{1-q}$.    Could someone please have a look at my working and show me where I have gone wrong? Or perhaps a neater calculation? Thanks!    PS I did check similar questions answered here, but couldn't find anything relevant to my problem.
55564
Let $X\sim B(n,p_h)$ and $Y\sim B(n,p_\ell)$ be two random variables following a respective binomial distribution, where $p_h&gt;p_\ell$. I want to show that  $$P(X\ge\alpha)\ge P(Y\ge\alpha),$$ for any $\alpha\in\{0,1,\dots,n\}$. In other words, I want to show  $$P(X\ge\alpha)=\sum_{i=\alpha}^{n}\binom{n}{i}p_h^i(1-p_h)^{n-i}\ge P(Y\ge\alpha)=\sum_{i=\alpha}^{n}\binom{n}{i}p_\ell^i(1-p_\ell)^{n-i}.$$  The statement to be proven is obviously true, but proving it is extremely difficult (at least for me). Can anyone give me some guidance on proving it?  It would be nice if someone can give an algebraic proof too, that is, show that  $$\sum_{i=\alpha}^{n}\binom{n}{i}p_h^i(1-p_h)^{n-i}\ge \sum_{i=\alpha}^{n}\binom{n}{i}p_\ell^i(1-p_\ell)^{n-i}.$$  Thank you very much!
67801
Let $Y_1, Y_2, ..., Y_n$ be iid random variables and $B_1, B_2, ..., B_n$ be Borel sets. It follows that  $P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)$...I think?  If so, does the converse hold true? My Stochastic Calculus professor says it does (or maybe misinterpreted him somehow?),  I was  the  that independence of the n random variables was equivalent to saying for any indices $i_1, i_2, ..., i_k$ $P(\bigcap_{j=i_1}^{i_k} (Y_j \in B_j)) = \Pi_{j=i_1}^{i_k} P(Y_j \in B_j)$.  So, if the RVs are independent, then we can choose $i_j=j$ and k=n to get $P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)$, but given $P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)$, I don't know how to conclude that for any indices $i_1, i_2, ..., i_n$ $P(\bigcap_{j=i_1}^{i_k} (Y_j \in B_j)) = \Pi_{j=i_1}^{i_k} P(Y_j \in B_j)$, if that's even the right definition.  p.17  seems to suggest otherwise. idk  Help please?  Also this:    or    So, this answer is to use the Omega part to establish pairwise independence and ultimately conclude independence. Without that assumption, we cannot conclude independence. Is that right? Why does that not contradict the definition of independence: $P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)$ ?
15671
Suppose  $X, Y$ are i.i.d and $X,Y\sim N(0,1)$. What is $P(X &gt; 0 \mid X + Y &gt; 0)?$   What I got so far is listed below:  $\begin{align*} P(X &gt; 0 \mid X + Y &gt; 0) &amp;= \frac{P(X+Y&gt;0\mid X&gt;0)\cdot P(X &gt; 0)}{P(X+Y &gt; 0)}\\ &amp;= P(X+Y &gt; 0\mid X &gt; 0)\\&amp;=P(Y &gt; -X\mid X&gt;0) \\&amp;=\int_{0}^{\infty}\int_{y=-x}^{\infty} \frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}dydx\end{align*}$  However, I don't know how to calculate this integral. Is there anything that I did wrong in the process? And also is there any easier way to approach this kind of problem?
38218
Let $X \sim Gamma(3,3)$ and $Y \sim Exp(1)$.  How do I calculate $P(X&gt;Y)$?  I believe I rewrite it as $P(X-Y&gt;0)$ but I am unsure how to calculate $X-Y$ for two different distributions?
25068
I am working on this problem, where I am required to write down the pgf of $X$, as well as the pgf of $Y$ given $X=j$, and then using conditional expectation find the pgf of $Z$.  So far, I have \begin{align*} G_{X}(z)&amp;=(pz+(1-p))^{n}\\ G_{Y|X=j}(z)&amp;=(qz+(1-q))^{n+j} \end{align*}  I'm new to generating functions and am having difficulty finding the pgf of $Z$. I thought I had a breakthrough when I wrote down \begin{align*} G_{Z}(z)&amp;=\mathbb{E}\left(z^{Z}\right)\\ &amp;=\mathbb{E}\left(z^{n+j-Y}\right)\\ &amp;=\frac{z^{n+j}}{G_{Y}(z)} \end{align*}  But I'm quite sure this isn't correct as I am then required to deduce that $Z$ has the same distribution as the sum of two independent Binomial random variables. I don't know how to apply conditional expectation here. I'm sure this is a very simple problem, and I just want to understand it. Thanks for your patience and time.
34581
I came across this question as an exercise, had a brief idea, but didn't know how to proceed.     Let $X \sim N(0, 1)$. What is the p.d.f of $|X|$ ?   I know the final p.d.f looks just like the right half of the original pdf, but extended vertically for a factor of 2.  Could someone please show mathematically what the p.d.f of variable |X| is going to be, and explain briefly? Thank you.
76179
Let X and Y have a trinomial distribution with join probability function:  $ f(x,y) = \frac{n!}{x!y!(n-x-y)!}p^x q^y (1-p-q)^{n-x-y}; x=0,1,...n, y=0,1,...n$  Let $T=X+Y$  I can't seem to figure out where to go exactly. Here's where I am at:  $\sum_{x=0}^{n} \frac{n!}{x!y!(n-x-y)!}p^x q^y (1-p-q)^{n-x-y}\mbox{ substitute in Y=T-X}$ $\sum_{x=0}^{n} \frac{n!}{x!(t-x)!(n-t)!}p^x q^{t-x} (1-p-q)^{n-t}$   I think I could I figure out the algebra.. if I could just see what to do with the summation. I know I have to involve a t somewhere, but I'm not sure. Maybe somebody could just show me how to set up my summation correctly with the substitution?
51403
Let $X\sim N(0,1)$ and let $Z$ be a random variable independent of $X$ such that:  \begin{equation} \Pr(Z=z) = \begin{cases} \frac{1}{2} &amp; \mbox{if $z = -1$ or $z=1$}, \\ \\ 0 &amp; \mbox{otherwise}. \end{cases} \end{equation}   Define $Y = ZX$. Show that $Y \sim N(0,1)$. Determine $\operatorname{Cor}(X,Y)$. Are $X$ and $Y$ independent?   Progress:  I'm basically stuck at the beginning. I got as far as $P(ZX &lt; y) = P(Z &lt; y/X)$, which I'm not even sure is correct. I'm also not clear on what it means by defining $Y = ZX$ - does it mean I need to state the probability density function or something? And how do I show that $Y \sim N(0,1)$?
87869
I have an answer but I'm not sure if I'm doing it right: $f(x)$ is the pdf $f(x)=2e^{-2x}$, then, if the random variable $Y=\ln(X)$, then we simply substitute:  $P(Y \leq y)$=  $P(2e^{-2X} \leq y)$  $P(-4X \leq y)$ since $y=ln(x)$  $P(X \leq - \dfrac {1} {4} y)$  This is the distribution function, right?  Thanks!
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!
91434
When I studied Bayesian statistics, a question about the notation of Bayes' Theorem came to my mind. Below is the density function version of Bayes' Theorem, where $y$ is data vector and $\theta$ is the parameter vector:  $$ p(\theta|y)=\frac{p(y|\theta)p(\theta)}{p(y)} $$ The numerator on the right handside can be written as: $$ p(y,\theta) $$ which is the joint probability distribution of $y$ and $\theta$, then Bayes' theorem could be written as: $$ p(\theta|y)=\frac{p(y,\theta)}{p(y)} $$ Furthermore, $$  p(\theta|y)\propto p(y,\theta) $$ Am I right on this? I think it does not look right. Because the posterior is the proportional to the joint density function. But where is the mistake?
60026
I realize that this question has been asked multiple times and I do not really want to know how to do it, I understand how to solve it, my issue is somehow I have used the distribution and I am not getting the right answer.  Method of Solving: I solved this in the first go by directly applying the distribution formula that is $$\binom {n+r-1}{r-1} $$ By using, $n = 10, r= 3$ we get, $$\binom{12}{2} = 66$$ however the answer given is $$\binom{12}{3} = 220$$ Next, I tried to solve by taking multiple cases that is, setting any variable, $x,y$ or $z$ to values from $0,1,2,3...10$, so for example if $x = 0$ then we get, $$y+z=10$$ now it is simple enough to distribute this even without the formula, we get $$\Bigl((0,10),(1,9),(2,8)(3,7),(4,6)\Bigl) \cdot 2,(5,5)$$ which gives us $11$ cases, that can be verified using the formula $$P_0 = \binom{11}{1}.$$ Now taking cases till $x=10$ we get, the final result as $$\sum _{n=0} ^{n=10} P_n = 11+ 10 + 9 + 8+ \ldots +1 = 66$$ Please tell me where I am going wrong or of it simply a case of the answer in the text being wrong, and if the second method is correct.
109702
I'm trying to get more familiar with $E(XY) = E(X)E(Y)$, and all examples that I have come across seems to satisfy this expression. My question then is, if there exist an example of two random variables $X$ and $Y$ for which $E(XY) \neq E(X)E(Y)$?  I am looking for an example using a joint probability distribution table.  Thank you.
3091
This is a follow-on from my  about samples from a distribution.  Suppose $X_1 \ldots X_{n-1}, X_n$ are random variables all following some fixed distribution $D$. How do I prove that $P(X_n &gt; X_1 \wedge \ldots \wedge X_n &gt; X_{n-1}) = \frac{1}{n}$ (this is true intuitively and experimentally).
25544
Say you take a dice and you roll it twice so that you have a pair (X,Y) where X represents the first roll, Y represents the second.   When you have a distribution like max(X,Y), what type of distribution is that? It's not uniform, obviously, because the outcomes of max(X,Y) are clearly more favorable towards the higher numbers. I'm being asked to "describe" the distribution. I know that we can look at each distinct outcome, like:  max(1,1) = 1  max(1,2) = 2  max(1,3) = 3  max(1,4) = 4  max(1,5) = 5  max(1,6) = 6  max(2,1) = 2  max(2,2) = 2  max(2,3) = 3  max(2,4) = 4  max(2,5) = 5  max(2,6) = 6  max(3,1) = 3  max(3,2) = 3  max(3,3) = 3  max(3,4) = 4  max(3,5) = 5  max(3,6) = 6  max(4,1) = 4  max(4,2) = 4  max(4,3) = 4  max(4,4) = 4  max(4,5) = 5  max(4,6) = 6  max(5,1) = 5  max(5,2) = 5  max(5,3) = 5  max(5,4) = 5  max(5,5) = 5  max(5,6) = 6  max(6,1) = 6  max(6,2) = 6  max(6,3) = 6  max(6,4) = 6  max(6,5) = 6  max(6,6) = 6  But is that the most effective way to look at a distribution like this? Is there a faster and more concise way to describe this distribution other than saying the probability of each outcome?  i.e.,  P(max = 1) = 1/36  P(max = 2) = 3/36 = 1/12  P(max = 3) = 5/36  P(max = 4) = 7/36  P(max = 5) = 9/36 = 1/4  P(max = 6) = 11/36  So, obviously, I understand how to LOOK at a discrete distribution and see what the probabilities are, but not until I look at each possible outcome. But what does it mean to "describe" a distribution? This isn't uniform or any sort of discrete distribution we're taught. Can someone provide clarity?