New answers tagged

1

You do not wish the $Y$ value to exceed the bounds of its support $(0;60)$. $$\begin{align}&{1\over 3600}\int^{60}_{0} \int ^{\min(60,x + 15)}_{\max(0,x - 15)}~\mathrm dy~\mathrm dx \\[2ex] & = {1\over 3600}\left(\int_0^{15}\int_{0}^{x+15} ~\mathrm dy~\mathrm dx +\int_{15}^{45}\int_{x-15}^{x+15} ~\mathrm dy~\mathrm ...


1

Regarding the calculus, I think it's wrong since you have three regions to consider for $y$: $0$ to $x+15$, $x-15$ to $x+15$, $x-15$ to $60$. But I don't claim to be good at calculus.... Alternatively, you can compute the complementary probability $$P(|Y-X|<15) = 1-2P(Y>X+15) = 1-2\left[\frac{\frac{1}{2}(60-15)(45)}{60^2}\right] = ...


0

Yes, you are correct. Sorry, both you and I were wrong. Should have paid attention to the r.v.'s support. Changes made in red. If you want to argue formally (rather than by drawing lines), you can use the law of total probability. You can do $$ \begin{align*} \Pr(-15 < X - Y < 15) &= \Pr(-15 + Y < X < 15 + Y) \\ &= ...


3

By the Borel-Cantelli lemma, if the series $$ \sum_{n=1}^\infty P\{|Y_n|>\varepsilon\} $$ converges for each $\varepsilon>0$, $Y_n\to0$ almost surely as $n\to\infty$. Using Chebyshev's inequality, $$ \sum_{n=1}^\infty P\{|Y_n|>\varepsilon\} \le\frac1{\varepsilon^2}\sum_{n=1}^\infty\operatorname E|Y_n|^2 $$ since $\operatorname EY_n=0$. By ...


2

Hint: use the Borel—Cantelli lemma. In more detail: fix any $\varepsilon > 0$, and let $A_n=A_n(\varepsilon)$ be the event $\{ \lvert Y_n\rvert > \varepsilon \}$. By Borel—Cantelli, to show that $Y_n \xrightarrow[n\to\infty]{\rm a.s.} 0$, it is sufficient to show $$ \sum_{n=1}^\infty \mathbb{P} A_n < \infty. $$ In even more detail: (place your ...


1

You gave a generic definition of the CDF of uniform continuous distributions. You have two specific uniform discrete distributions.   These will have step functions. The CDF of the minimum of those RV will also be a step function, although the steps will not be uniform .   We also know the support of $Z$ since: $\min\{X, Y\}\in \{0,1,2\}$ ...


1

Hint: If $X$ takes on two possible values and $Y$ takes on three possible values, then the pair $(X,Y)$ has six possible values, each one equally likely (since $X$ and $Y$ are independent and each one has uniform distribution). This means you can enumerate the possible values for $Z:=\min(X,Y)$ and determine their probabilities (construct a table!). You'll ...


0

I think you just forgot the factor $\frac{3}{2}$; you did not compute the unconditional probability of $X = 2$. We are told that $X|U\sim\text{Geom}(U)$, and that $U\sim\text{unif}[1/3,1]$. Then \begin{align*} P(X=2)&=\int_{1/3}^3P(X=2|U=u)\cdot f_U(u)\,du\\ &=\int_{1/3}^1(1-u)^{2-1}u\cdot\frac{1}{2/3}\,du \\ &= ...


1

A possibility for (i): define the random variable (indicator) $X=\mathbf{1}_{N=0}$, so that $ \mathbb{P}\{N=0\} = \mathbb{E} X $. Now, write $$ \mathbb{E} X = \mathbb{E}[\mathbb{E}[ X\mid \Lambda ]] $$ so we can first deal with the random variable $$ \mathbb{E}[ X\mid \Lambda ] = e^{-\Lambda} $$ and then get $$ \mathbb{E} X = \mathbb{E} e^{-\Lambda} = ...


1

Since $N\mid\Lambda=\lambda \;\sim\;\mathcal{Pois}(\lambda)$ you know: $$\mathsf P(N=0\mid \Lambda=\lambda) = \dfrac{\lambda^0 \mathsf e^{-\lambda}}{0!}$$ Since $\Lambda\sim\mathcal{U}(0;5)$ you also know $f_\Lambda(\lambda) = \frac 1 5 \mathbf 1_{\lambda\in(0;5)}$ And you should know how to find marginal distributions. $$\mathsf P(N=0) =\int_0^5\mathsf ...


1

Yes, all your previous calculations look fine. Recall that to find the unconditional distribution of $N$ can be computed by $$P(N = k) = \int_0^5 P(N=k\mid\Lambda = \lambda)f_\Lambda(\lambda)\,d\lambda.$$ Notice that you are told $N$, and so you are seeking the conditional distribution of $\Lambda$ given $N$, which is $$f_{\Lambda|N}(\lambda\mid k) = ...


0

a) It is given that $X_{|U=p} \sim Geo(p)$, so $E[X|U=p] = \frac{1}{p}$ and $Var[X|U=p] = \frac{1-p}{p^2}$. b) Using the total expectation and total variance: $$ E[X] = E[E[X|U]] = E\left[\frac{1}{u} \right] = \int_{1/3}^1\frac{1}{u}\frac{3}{2}du = \frac{3}{2}(0 + \ln(3)) \approx 1.65 $$ For the variance of $X$ use the total variance formula, i.e., ...


2

There are $32^4$ possible outcomes for $(a,b,c,d)$, since each one has $32$ outcomes. a) If none of them are equal, then outcomes for $(a,b,c,d)$ are $32\times 31\times 30\times 29$. Thus, the probability is $\frac{32\times 31\times 30\times 29}{32^4}=\frac{13485}{16384}=0.823$ b) Select the common number: $32$ outcomes. Select the places: $\binom{4}{2}$. ...


1

Consider a sequence of four observations, one from each random variable: $$(x_1,x_2,x_3,x_4).$$ There are $32^4 = 1{,}048{,}576$ such sequences, each of which is equally likely. For a), there are 32 choices for $x_1$, leaving 31 choices for $x_2$, leaving 30 choices for $x_3$, leaving 29 choices for $x_4$. By the multiplication principle, there are $32 ...


0

Hint: Let $B$ be the time (in minutes) elapsed since 10 o'clock. Then $B$ is uniformly distribute on the interval $[0,30]$ with density $\dfrac{1}{30}$. The probability that the waiting time is longer than 10 minutes is $P(B ??)$. If the bus hasn't arrived at 10:15 (we're given a condition, so think of the definition of conditional probability), the ...


0

You want to show that $T_x\mid T>x$ is uniformly distributed on $(0,b-x)$. One way to do this "using consistent notation with probability theory", is to calculate the cdf of this random variable, i.e. of $T_x \mid T>x$ in terms of the known cdf $F_T$ of the random variable $T$. You know how the cdf should look (you are correctly intuitively guessing ...


2

I'll try not to give away the entire answer, but as you said it does seem pretty obvious. The issue is then converting this into rigorous mathematics! Let's proceed as follows. We're interested in $P(T_x \le t)$, for varying $t$. Now, in terms of events, $$ \{ T_x \le t \} \Leftrightarrow \{ T \le t + x \},$$ by definition of $T_x$, and so we know that ...


1

If you were given\taught the Beta function, then, since you know the density of $M$ (the minimum), \begin{align*} E[M] &= \int_0^1 mf_M(m)\,dm\\ &=\int_0^1m\cdot n(1-m)^{n-1}\,dm\\ &=n \int_0^1\cdot m^{2-1}(1-m)^{n-1}\,dm\\ &=n B(2,n)\\ &=n \cdot\frac{(2-1)!(n-1)!}{(2+n-1)!}\\ &=\frac{n!}{(n+1)!}\\ &=\frac{1}{n+1}. \end{align*} ...


1

Here I gave you explicitely forms of each distribution. For $N$: $\begin{eqnarray}E[N]&=&\int_0^1nx(1-x)^{n-1}dx\\ &=&\frac{n}{n^2+n}\\ &=&\frac{1}{n+1} \end{eqnarray}$ $\begin{eqnarray}E[N^2]&=&\int_0^1nx^2(1-x)^{n-1}dx\\ &=&\frac{2n}{n(n+1)(n+2)}\\ &=&\frac{2}{(n+1)(n+2)} \end{eqnarray}$ Can you do for ...


0

A general aproach is: $\begin{eqnarray} F_M(m)&=&P(M\le m)\\ &=&P(X_1\le m,...,X_n\le m)\\ &=&P(X_1\le m)...P(X_n\le m)\\ &=&P(X_1\le m)^n\\ &=&F_X(m)^n \end{eqnarray}$ Thus $f_M(m)=nF_X(m)^{n-1}f_X(m)$. In the same spirit, $F_N(m)=1-(1-F_X(m))^n$. and $f_N(m)=n(1-F_X(m))^{n-1}f_X(m)$. On the other hand, suppose ...


0

Not quite. Notice that $$P(M\in dm,N\in dn) = P(X_{(1)} \in dm,X_{(n)}\in dn)$$ For a moment, let's pretend that $X_1 \in dm$ and $X_n\in dn$. This forces $X_2,\dotsc,X_{n-1}\in (m-n)$. This equates to \begin{align*} P(X\in dm) &= f_{X_1}(m)\,dm = 1\,dm\\ P(X\in dn) &= f_{X_n}(n)\,dn = 1\,dn\\ P(m< X_k<n) &=\frac{n-m}{1} = n-m ...


0

You want $f_{X_{(n)},X_{(1)}}(M,N)$ To hit the favoured space: one of the $n$ variables needs to be the maximum value $M$, with density $f_X(M)$. given that, then one of the remaining $n-1$ variables then needs to be the minimum value $N$ with density $f_X(N)$. given that, then each of the remaining $n-2$ variables to lie between $(N;M)$, and each of ...


1

Consider the events of any of the $n$ variables being maximum.   Because there must be a maximum, these $n$ events are exhaustive.   Because they are continuous, then ties are nigh impossible, so these events must be disjoint.   Because they are identical and independently distributed, these events are equally probable.   Thus one, and ...


0

Note that $\begin{eqnarray} 1&=&P(M=U_1\cup M=U_2\cup...\cup M=U_n)\\ &=&P(M=U_1\cup M=U_2\cup...\cup M=U_n|U_i\neq U_j\,\forall i,j)P(U_i\neq U_j\,\forall i,j)\\ &&+P(M=U_1\cup M=U_2\cup...\cup M=U_n|\exists i,j\,s.t.\,U_i= U_j\,)P(\exists i,j\,s.t.\,U_i= U_j\,)\\ &=&P(M=U_1)+P(M=U_2)+...+P(M=U_n)\\ &=&nP(M=U_1) ...


1

The fact that there must be a duplicate first name/last name pair means there are at least ???? people in the population. For how big a population would $225,000$ be $7.4\%$? For how big a population would $225,000$ be $8.6\%$? I think you are supposed to find the interval of population that meets the criteria, assume that all values are equally likely ...


0

If you generate the points uniformly on the unit circle, you will get the same distribution as if you generate them uniformly on the upper half circle. The fraction of points in the interval $[a,b]$ will be the fraction of the arc length between $x=a$ and $x=b$. The angle at $x=a$ is $\arccos a$ and at $x=b$ is $\arccos b$, so the fraction in $[a,b]$ is ...


0

Here is an illustration using simulation of the case in which the points are uniformly distributed within the unit disk. I generate 50,000 points in the square with vertices at $(-1,-1)$ and $(1,1)$ and discard those outside the circle. Then look at a histogram of the $x$-coordinates of the remaining 39,236 points (within the circle), which gives a pretty ...


3

If the distribution is discrete, there may be a problem. For example, imagine tossing a fair coin that has a $0$ on one side and a $1$ on the other. Then all possibilities $(0,0)$, $(0,1)$, $(1,0)$, and $(1,1)$ are equally likely. When we order, the result $(0,1)$ is twice as likely as either $(0,0)$ or $(1,1)$. Remark: If the underlying distribution is ...


1

A formal argument is: Let $(\Omega,\mathcal{F},P)$ a probabilistic space and $A:\Omega\to\mathbb{R}$ a random variable. Set $g:\mathbb{R}\to\mathbb{R}$ such that Borel measurable. Then $g\circ A:\Omega\to\mathbb{R}$ is random variable. Now, let $U$ a Borel set in $\mathbb{R}$. Suppose $U\subseteq(Im(g))^c$ (that is, $U$ doesn't have elements of $Im(g)$). ...


1

Assume independent uniform distribution of the car and truck arrival, and using time indexing from minute $0$ to $15$ (instead of $7:15 - 7:30$). For $A$ the event of arrival (of either car or truck) we have that $A$ ~ $ Uniform(0,15)$. Because it will be enough if car and truck arrive between $5$ and $7$ (between $7:18$ and $7:22$), It follows that (1) ...


0

We know that $P(X_i = X_j) = 0$ so we can say, \begin{equation} \tag{1} P\left(X_1 < X_2 < X_3\right) = P\left(X_1 \le X_2 \le X_3\right) \end{equation} Thus, the problem is the symmetry group $S_3$ and you are simply asking for the probability of a particular permutation of $X_i$. The number of permutations is simply the binomial coefficient, ...


1

You were close, but you just needed to integrate the variable over its support interval. $$\begin{align}\mathsf P(X_1<X_2<X_3) & = \int_0^1 f_{X_2}(x_2)\; F_{X_1}(x_2)\; (1-F_{X_3}(x_2))\operatorname d x_2 \\[1ex] & = \int_0^1 1\cdot x_2\cdot(1-x_2)\operatorname d x_2 \\[1ex] & = {\left[\tfrac 1 2 {x_2}^2-\tfrac 1 3{x_2}^3 ...


2

Each of the three values $X_1, X_2, X_3$ has an equal chance of being the middle value.


5

You can solve this problem by symmetry. The following events are distinct $i$) $X_1$ lies between $X_2$ and $X_3$ $ii$) $X_2$ lies between $X_1$ and $X_3$ $iii$) $X_3$ lies between $X_1$ and $X_2$ and they are complete in the sense that one of them is always true, $P(i) + P(ii) + P(iii) =1$. Because of the symmetry they are equivalent and we have $P(i) ...


1

You are partially correct. The probability of three 2's is $\left(\frac16\right)^3=\frac{1}{216}$. The probability of all three the same is $1\times\left(\frac16\right)^2=\frac{1}{36}$ as the first number can be any value. The probability of two the same and the third different is: $1\times\frac{1}{6}\times\frac{5}{6}=\frac{5}{12}$. Same argument applies ...


1

You're right, the probability of all 2s is $$P(222) = P(2)P(2)P(2) = \frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{6} = \left(\frac{1}{6}\right)^3= \frac{1}{216},$$ where use the product by independence. When calculating probability that they are all the same is, you have to account for all the choices you have for a match, which is $\binom{6}{1}$. So, ...


0

I calculated using this way the blue shaded area is$\frac{9}{4}$and the total area is $4$ . Hence the probability will be $\frac{9}{4}*\frac{1}{4}$


1

Assuming they are independent, let $M = \max\{X,Y\}$. Then, \begin{align*} P(M\leq 1/2) &= P(X\leq 1/2, Y\leq 1/2)\tag 1\\ &= P(X\leq 1/2)P(Y\leq 1/2)\tag 2\\ &=\left(\frac{1/2-(-1)}{1-(-1)}\right)\left(\frac{1/2-(-1)}{1-(-1)}\right)\tag 3\\ &=\left(\frac{3/2}{2}\right)^2\\ &=\frac{9}{16} \end{align*} where: $(1)$ is true since if I ...


0

I'm not entirely sure what you are doing, but I think your bounds are wrong. It might be $0\leq s_1 \leq s_2$. \begin{align} P\left.\left[S_2\le\frac12\right|N(1)=2\right] &= P\left.\left[S_1\le\frac12, S_2\le\frac12\right|N(1)=2\right]\\ &=\int_{0}^{\frac12}\int_{0}^{s_2} f_{S_1, S_2|N(1)=2} (s_1, s_2 | 2) {ds_1}{ds_2}\\ ...



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