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1

The random variable $Z$ of your question has density function $e^{-z/2}$ for $z\gt 0$ and $0$ for $z\lt 0$. By your argument, we have $$\Pr(Y\le y)=\Pr(e^Z\le y).$$ If $y\le 1$, then $\Pr(e^Z\le y)=0$, since $Z\le 0$ with probability $0$. Thus the density function of $Y$ is $0$ for $y\le 1$. If $y\gt 1$, then $$\Pr(e^Z\le y)=\Pr(Z\le \ln y)=1-e^{-(\ln ...


0

Please read Darren Wilkinson's blogpost Getting started with parallel MCMC. Getting a deterministic computer to generate a pseudo-random number stream is nontrivial, and there is no reason to think streams starting from different seeds (as in the other question you referenced) will be independent, given that this was not a desideratum when designing the ...


0

To do a scientific experiment truly objectively, the ideal process is: Describe the experiment to be performed. Enumerate all possible outcomes of the experiment. Declare what conclusion you would draw from each outcome. Perform the experiment. Publish the results and the conclusion. People often simplify this process in ways that can introduce minor (or ...


0

I should point out that testing the hypothesis that the coin is fair is most definitely not equivalent to testing the hypothesis that the sequence of heads/tails it generates is random. The sequence can be random but biased; conversely, the coin can be fair but in a deterministic (non-random) fashion.


0

A $p$-value is actually defined (in Statistical Inference by Casella and Berger, for instance) to be a statistic, not just a number, and it is called valid if it satisfies $$ P_\theta[p({\bf X}) \leq \alpha] \leq \alpha $$ for $\theta \in \Theta_0$. So the probability of seeing small values is genuinely low under the null hypothesis. This makes sense. ...


2

The best you can hope for is $\left\lceil\operatorname{lg}(152)\right\rceil=8$ guesses in the worst case. If you are limited to 6 questions, you won't be able to guarantee a correct answer, in which case the game becomes maximizing your chance of winning. You could, for example, split off the 32 most likely cases (accounting for 161 of the 400 possible ...


0

You have to decide whether you want the best chance to get a unique value, or the worst case number of possibilities. There are $152$ unique products that can be produced. As $2^6=64$, in the worst case you can guarantee getting down to $2$ or $3$ numbers after $6$ questions. For a reasonable shot at finding a unique number, ask first if it is less than ...


3

Using theorem of transformation variables: $$ g_Z(z)=f_Y(y)\ |J|=f_Y(y)\ \left|\frac{dy}{dz}\right|, $$ where $J$ is Jacobian. We have $$ f_Y(y)=\lambda e^{-\lambda y}=e^{- y}\quad;\text{ since $\lambda=1$ and for }y\ge0 $$ and $y=-z$. Therefore $$ g_Z(z)=e^{- y} \left|\frac{dy}{dz}\right|=e^{-(-z)} \left|\frac{d}{dz}(-z)\right|=e^z\ ...


2

We want $f_Z(z)$, the density function of $z$. We will first look for the cdf of $Z$. When $z\gt 0$, we have $\Pr(Z\le z)=\Pr(Y\ge -z)=1$. So for $z\gt 0$, we have $F_z(z)=1$, and therefore $f_Z(z)=0$. Now let $z\lt 0$. Then $$F_Z(z)=\Pr(Z\le z)=\Pr(Y\ge -z)=1-F_Y(-z).$$ Differentiate with respect to $z$, using the Chain Rule, and the fact that the ...


2

Yes, uncorrelated means that their covariance is zero. Furthermore, if two random variables are independent, then they are indeed uncorrelated. However, if they are uncorrelated, that does not mean that they are independent. i would advise reading http://en.wikipedia.org/wiki/Uncorrelated


2

The setup is correct, we want $$\Pr(X=0)\Pr(Y=0)+\Pr(X=1)\Pr(Y=1)+\Pr(X=2)\Pr(Y=2).$$ We have $\Pr(X=0)=(0.7)^3$ and $\Pr(Y=0)=(0.4)^2$. For $\Pr(X=1)$, the right expression is $\binom{3}{1}(0.3)(0.7)^2$. Similarly, $\Pr(Y=1)=\binom{2}{1}(0.6)(0.4)$. Finally, $\Pr(X=2)=\binom{3}{2}(0.3)^2(0.7)$ and $\Pr(Y=2)=(0.6)^2$. Remark: There is nothing ...


0

Atleast one of the three colors not present = $P(E)= 1-P(All) = 1-P(A \cap B \cap C)$ Now use your formula the other way round.


0

Whenever you have a function $f \,:\, A \to B$ and some $\sigma$-Field $\mathcal{B} \subset \mathcal{P}(B)$ on $B$, then the set of all preimages of sets in $\mathcal{B}$, i.e. $$ \mathcal{A_f} := \left\{ f^{-1}(Y) \,:\; Y \in \mathcal{B}\right\} $$ is a $\sigma$-field on $A$. Note the use of $f^{-1}$ here doesn't imply that $f$ is invertible - $f^{-1}(Y)$ ...


1

Draw a picture. We will use it to set up the integration. We could also treat the problem as a pure geometry problem. We do it in somewhat greater generality, so that you can deal with non-constant joint densities. For $f_X(x)$, we want to "integrate out" $y$. Let $-1\le x\lt 0$. Then $y$ goes from $0$ up to the line that joins $(-1,0)$ and $(0,1)$. That ...


2

The unconditional probability that $X-Y=0$ is not $1$. The conditional probability that $X-Y=0$, given that $X+Y=0$, is obviously $1$. So our random variables are not independent. Alternately, and almost equivalently, we have $\Pr((X+Y=0)\land (X-Y=0))=\Pr(X=0)\Pr(Y=0)=e^{-2}$. But $\Pr(X+Y=0)=e^{-2}$ and $\Pr(X-Y=0)\lt 1$, and therefore ...


0

Create an indicator variable for $XY$. The possible values of $XY$ are 0 and 1. So $$XY = \begin{cases} 1 \ \ \text{if} \ X \ \text{and} \ Y \ \text{occur} \\ 0 \ \ \text{if} \ \text{only} \ X \ \text{occurs}, \ \text{only} \ Y \ \text{occurs} \ \text{or neither occurs} \end{cases}$$


1

You can rewrite $(A\cup B)\cap (C\cup B)=(A\cap C)\cup B$, and then \begin{align*} \mathbb{P}\left((A\cap C)\cup B\right) &= \mathbb{P}\left(A\cap C\right) + \mathbb{P}B - \mathbb{P}\left(A\cap C\cap B\right) \\ &= \mathbb{P} A\cdot \mathbb{P}C + \mathbb{P}B - \mathbb{P} A\cdot \mathbb{P}C\cdot \mathbb{P}B \\ &= \frac{1}{4} + \frac{1}{2} - ...


1

Parts $1$ and $2$ are very easy: If you want an $m \times n$ rank $r$ matrix, then compute $UPV$ where $U$ is $m \times m$ invertible $V$ is $n \times n$ invertible and $P$ the $m \times n$ matrix that looks like $$P = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix}$$ with $r$ ones. ...


1

What you can conclude it is the events $$\{X+Y+Z=1\} \cap \{X=Y\} = \{2X+Z=1\} \cap \{X=Y\}$$ $\cap$ means AND. Note that the event $\{X+Y+Z=1\} \cap \{X=Y\} \neq \{2X+Z=1\}$ in general. For example, let be $w$ such that $X(w)=0$ and $Y(w)=Z(w)=1$. Such $w$ is a possibility in $\{2X+Z=1\}$ but it is not in $\{X+Y+Z=1\} \cap \{X=Y\}$. Then the events ...


1

$P([X+Y+Z=1]\,and [X-Y=0]) = P(X=0\,and\,Y=0\,and\,Z=1)= (\frac{1}{2})^3$


1

If you look at the accepted answer for the post you linked, you see that the direction is generated by sampling from a multivariate normal distribution, and then the radius is chosen according to the desired distribution. In your case, you don't need to define a minimal radius $R_0 > 0$ because when you multiply the scaling factor $1/r$ for probability of ...


0

This type of questions is often asked on job interviews, so I share a full solution of this more general form. Q: We have 3 variables, $X, Y, Z$. In what range can be the correlation between them? A: We can put the 3 variables in a correlation matrix \begin{equation*} \left( \begin{array}{c c c} 1 & \rho(X, Y) & \rho(X, Z) \\ \rho(Y, X) & 1 ...


1

You need a final term that give the probability of unlocking with the $k$th key. So $P(X=k) = (1-\frac{1}{N})(1-\frac{1}{N-1})\cdots(1-\frac{1}{N-k+1})\frac{1}{N-k}$ $=\frac{N-1}{N}\frac{N-2}{N-1}\cdots\frac{N-k}{N-k+1}\frac{1}{N-k} = \frac{1}{N} $ by cancellation. The approach suggested by the solution in the question is combinatorial rather than ...


0

This is an asymptotic result assuming independence of trials. Use CLT to show that $\sqrt n((\frac{1}{n}\sum X_i)-p)=O_p(1)$. So, $\bar S_n=\frac{S_n}{n}$ is $\sqrt n$ bounded. Hence, if you change the rate of convergence slightly, it becomes unbounded (your limit goes to 1) or converges in probability to 0 (in this case you get probability 0 in the limit). ...


1

The reality is that few coins are actually truly 50/50. If the center of gravity of the coin does not lie perfectly between the landing surfaces of the two sides then there will be some slight bias toward one side. Differences in the images imprinted on the two sides of a traditional coin normally cause the center of gravity to be closer to one side or the ...


0

If there was no correlation between your (conscious and unconscious) knowledge of the SD card (which includes knowledge of any previous flips) and your choice, then it would be fair. However, this will not be the case. The unknowing of your own psychology is not enough to eliminate the bias, no matter if the bias is negative or positive.


0

Both of the other answers hinted at this: How do we choose a number "at random" in the unit interval [0,1)? The problem is that we would have to specify infinitely many digits. One standard way to "do this" is as follows: Pick whether it is in the first half of the interval or the second half; define each to have probability 1/2. We have just picked the ...


0

The process of picking out a number from $0$ to $1$ from a continuous distribution involves picking a number from smaller and smaller sub-intervals because the probability of picking any one number goes to zero as the size of the interval becomes arbitrarily small. The set of rationals can be covered by a set of intervals of arbitrarily small total length ...


0

You need to know how the power spectrum is transformed when the process is filtered by a linear time-invariant filter. Let $Y(t)$ be the output of a linear time-invariant filter with frequency response $H(f)$, and $X(t)$ is the input signal. Then the power spectral density of the output process $Y(t)$ is given by $$S_Y(f)=S_X(f)|H(f)|^2$$ What remains is ...


4

I would say that the SD card itself is not a 'fair coin', since a reasonable requirement of a 'fair coin' would be that the probability of landing on each side is essentially 0.5 each. However, I think the entire process as a whole of assigning an outcome to each side of the card and then flipping the card is a 'fair coin': The reason why we need coins to ...


0

As an aside, related to the question but not actually answering it: even if you assign sides fairly and randomly, and keep the assigned sides after the first toss, you still can't do multiple flips and still get the same outcome as a fair coin: Suppose the probability of a "head" is $p$, and the probability of assigning event $A$ to heads is $1/2$. Then: ...


0

It COULD be a fair toss. Prior knowledge of anything but adequate testing is of no help to you. You just cannot establish fairness without testing. It MAY be determined to be fair in the future - AFTER testing. Not knowing doesnt make it fair. Any deviation from determining fairness by testing is pointless. You could only convince someone who didnt ...


1

There are various models of probability that mathematicians may use, such as epistemic, frequentist, or the trivial deterministic model where we assume the future is fixed. From a pure maths point of view picking any of those models is fine, so long as you don't mix models. For example, I really don't know what the 99th digit of $\pi$ is. So I can say that ...


0

you have a chance of $p_1$ that you choose a certain side, when you flip it there is a chance $p_2$ that that side comes up this means that you have a chance of $p_1*p_2+(1-p_1)*(1-p_2) = 2*p_1*p_2-p_1-p_2+1$ to win and $p_1*(1-p_2)+(1-p_1)*p_2=p_1+p_2-2*p_1*p_2$ that you lose. if neither is a perfect 50% (you are biased to picking a side and the card is ...


0

The number of toothbrushes that are not defective is: $325 - 14 = 311$. So the probability that a randomly selected new toothbrush is not defective is: $\dfrac{311}{325} = 0.957$


1

If you toss the SD card, and hide the result. Then, not knowing anything (nor the result, nor the probability of each side), you pick one side. We can then assume you had a 50% chance of picking a side — as you can not distinguish which side has greater probability. You then uncover the SD card. This way, the probability of each side is not taken into ...


1

"If I'm just as likely to assign an outcome to one side of this SD card as to the other, then it must be a fair." I think you're right with the reasoning. However, the SD card has a definite "front/back", which might be associated with "right/wrong" or "good/bad" in your head. I don't know how strong your bias is, but I would tend to assign the thing I ...


4

Yes, it's fair for the first time you use it, then it won't be fair anymore. If you don't know the possible outcome of something, and randomly map the outcomes of the thing to some result, the first time you use it it'll be fair (since the outcome was unknown by you). After the first time you use it, you've learnt something from it, and that can bias your ...


1

As has been pointed out, there are 2 events involved: the decision as to which side of the SD card to associate with which course of action and the flipping event which determines which side lands upwards. It's normal for those two events to occur in that order. But not necessary: an umpire may toss a coin and conceal the result while the opposing captains ...


1

TL;DR: carry a true random number generator with known distribution on your person at all times[*]. If there's no correlation between your assignment of outcomes to faces of the SD card, and the SD card's bias, then yes, it doesn't matter whether the SD card is biassed. Saying "no correlation" means it's as if the outcomes were assigned to the faces of the ...


1

Why have we over-complicated this? Assuming neither of you knows the bias that the SD card has as to which side it will most likely land on, then you both have an equal chance to pick either "heads or tails," therefore making the toss fair...


1

I feel there are some disturbing implications if the above were true, but I'm unable to convince myself it definitely isn't true. Why? Is it because you think that the SD card is obviously not balanced, so it cannot possibly be fair? But you said: Given that I don't know the bias of this SD card, would flipping it be considered a "fair toss"? If ...


0

I'll give it another try: Assume that the SD card falls on the back with probability $p$ and on the front with probability $1-p$. Further assume that you decide for option $A$ if the SD card falls on its back with probability $q$, and for $A$ with probability $1-q$ if the card falls on its front, and for option $B$ in the other cases. Then $$ P(A) = pq + ...


8

My look at it is this. You are moving probability from one place to the other. Assume side 1 has probability $p$, and side 2 $1-p$. If you were to use a fair coin to determine which side of the SD card you get, then you end up with $0.5p + 0.5(1-p)$ of flipping the side you picked(dictated by the coin). This happens to be exactly $0.5$. EDIT: As this ...


46

That is a very good question! There are (at least) two different ways to define probability: as a measure of frequencies, and as a measure of (subjective) knowlegde of the result. The frequentist definition would be: the probability of the sd card landing "heads" is the proportion of times it lands "heads", if you toss it many times (details ommited ...


10

You could say that there are two unknowns when you flip something: the probability to get side $A$ and the actual result. So, say you have $3$ coins and $p$ is the probability you get heads: Coin $A$ is fair ($p=0.5$) Coin $B$ always gives heads ($p=1$) Coin $C$ always gives tails ($p=0$) Obviously, the only fair coin is $A$. But if you don't know which ...


5

Imagine taking into account a "fair" coin's initial angular momentum, its initial velocity, initial orientation, the nearby air movements, etc. Assume we can disregard quantum effects and solve for the rest position of the coin in terms of purely classical physics. Assume you'll immediately find and correct any mistakes in your computation. You should ...


2

If the conceptual goal of a fair coin is to provide equal chances of two outcomes, then the problem you face is twofold: Is the SD biased? Can you fairly assign the real-world outcomes to the sides of the SD card? My hunch is that (1) doesn't matter as long as you can achieve (2), but (2) essentially requires a "fair coin", which you don't have.


4

The whole idea of a biased coin is basically a convenient fiction useful for making probability examples. It's not really possible to make an object with a coin like aspect ratio significantly biased when tossed. It can be biased when spun (EDIT: or I guess if allowed to bounce). So yes it's actually very close to fair, not just due to your ignorance of it's ...


195

Here's a pragmatic answer from an engineer. You can always get a fair 50/50 outcome with any "coin" (or SD card, or what have you), without having to know whether it is biased, or how biased it is: Flip the coin twice. If you get $HH$ or $TT$, discard the trial and repeat. If you get $HT$, decide $H$. If you get $TH$, decide $T$. The only conditions ...



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