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12 votes
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How many rolls are sufficient to ensure, with probability 99%, that the sum is greater than 100?

Let $X_1,...,X_k$ be i.i.d. Uniform($\{1,...,m\}).$ Then the distribution of $S_k=X_1+...+X_k$ is given exactly by a recursion proved in this paper$^\dagger$, which expresses this distribution in ...
r.e.s.'s user avatar
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8 votes
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Sum of iid random variables

Use "pgf"s (Probability generating functions). Here, any $X_i$ has the following pgf : $$g(s)=\tfrac19(4+4s+s^2)$$ It is essential to note that $$g(s)=\tfrac19(s+2)^2\tag{1}$$ The pgf of a ...
Jean Marie's user avatar
6 votes
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Independent random variables with $X^2 + Y^2 =1$

Example ... $X$ and $Y$ are independent, and both have the scaled Radermacher distribution $$ P(X=1/\sqrt2) = P(X=-1/\sqrt2) = 1/2. $$ Then $X$ and $Y$ are not constant, they are independent, but ...
GEdgar's user avatar
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6 votes

How many rolls are sufficient to ensure, with probability 99%, that the sum is greater than 100?

While not a rigorous answer, I don't know how to get the number analytically, this should suffice. Let $X_i $ be the result of the ith roll and $X_{i}\in\left\{ 1,2,3,4,5,6\right\} $ with equal ...
Danny Blozrov's user avatar
4 votes

How many rolls are sufficient to ensure, with probability 99%, that the sum is greater than 100?

One way to do this is to invoke the central limit theorem which says that as you sum more and more i.i.d random variables, their sum get close to a Gaussian. Let's assume that by the time the sum has ...
Rohit Pandey's user avatar
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3 votes

Independent random variables with $X^2 + Y^2 =1$

If $X$ and $Y$ are independent and $X^2+Y^2=1$, then $X^2$ and $Y^2$ are almost surely constant. Indeed, let $U=X^2$ and $V=Y^2$. Then $U=1-V$ and the random variables $U$ and $V$ are independent ...
Davide Giraudo's user avatar
3 votes
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Calculating Density Function and Conditional Expectation of Independent Exponential Random Variables

A note on how to "derive" the convolution formula (it is not the most general version). Let $X,Y$ be independent random variables with continuous densities $f_X, f_Y$. Using conditional ...
Peter Strouvelle's user avatar
3 votes
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Understanding the use of Indicator function

The event $D\leq \frac{c}{2}(T_c-\frac{cT_c}{2B_s})$ does not involve any random variable. So you can prove $A=B$ by considering the cases $D\leq \frac{c}{2}(T_c-\frac{cT_c}{2B_s})$ and $D> \frac{...
geetha290krm's user avatar
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3 votes

How should independent random variables be distributed to have their product be distributed as a gaussian?

If $X$ has a normal distribution with mean $0$, we can write $X = S \exp(L)$ where $S = \pm 1$, each with probability $1/2$, and $S$ and $L$ are independent. The product of $n$ iid random variables $...
Robert Israel's user avatar
2 votes

Simple question on the meaning of $y$ in $Y = y$

In probability theory we are working with a probability space, that is a triplet $(\Omega, \mathcal{F}, \mathbb{P})$ consisting of: $\Omega$ a sample space. This is a set of all possible outcomes of ...
Josef K.'s user avatar
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2 votes
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Show $E[Y | E[Y | X]] = E[Y | X]$.

@DagunDagun gave great hints. I will write it out more explicitly for my own benefit. Let $\sigma(X)$ denote the $\sigma$-algebra generated by the random variable $X$. It would suffice to show $\...
msantama's user avatar
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2 votes

Show $E[Y | E[Y | X]] = E[Y | X]$.

You should be done using the definition of the conditional expectation. To be more precise, recall that on a probability space $(\Omega, \cal F, \mathbb{P})$, given a random variable $X$ and a $\sigma-...
DagunDagun's user avatar
2 votes
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Can't understand expected gain calculation in first price auction

In the statement OP mentioned, $x$ is seen as fixed and $b$ is seen as a random variable (note the sentence "A does not know the value of $b$"). Hence, $x \geq f(b)$ happens with probability ...
Vezen BU's user avatar
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2 votes
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Is every collection of discrete random variables a function of independent random variables?

Yes, this is always possible. One method would be to take $N = 2^n-1$, and for each $X^i = (X^i_1,X^i_2,\cdots,X^i_n) \in \Omega$ let $p_i := \mu(\{X^i\})$. Let \begin{align*} Y_1 &\sim \text{...
user6247850's user avatar
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2 votes
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If $X\sim N(\mu, \sigma^2)$ and $\Phi$ is the CDF of a standard Normal random variable, what is the distribution of $\Phi(X)$?

Since $\ \Phi\ $ is strictly increasing, it has a strictly increasing inverse $\ \Phi^{-1}:(0,1)\rightarrow(-\infty,\infty)\ .$ Then, for $\ 0<x<1\ ,$ \begin{align} P(\Phi(X)\le x)&=P\big(X\...
lonza leggiera's user avatar
1 vote

Sum of iid random variables

@ABlack's assumption that there should be probabilities in the final sum is correct. Here we show that @ABlack's approach is correct and leads to the same result as @JeanMarie's elegant answer. In the ...
Markus Scheuer's user avatar
1 vote

discrete random variable probability question

Big thanks for @lulu for the help! the answer to this question is $\frac{\binom{8}{5} \binom{17}{10}}{\binom{25}{15}} = 0.3332$ in my understanding we choose 5 water bottles from the group of 8 using ...
Ellie's user avatar
  • 111
1 vote

What probability distribution function is this?

This is just my (not very rigorous) summary of the arguments in the comments (community wiki). Let $R$ be the number of runs ($R=3$ in your example). In the original setting we have $L$ as a fixed ...
1 vote
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An upperbound for the variance of mean estimator in higher dimensions

For $i \neq j$, as $X_i, X_j$ are independent and identically distributed, we have: $$ \mathbb{E}[\lVert X_i - X_j \rVert_2^2] = \mathbb{E}[\lVert X_i - \mu \rVert_2^2] + \mathbb{E}[\lVert X_j - \...
Thành Nguyễn's user avatar
1 vote

Distribution of difference of two random variables

The key bit from the Wikipedia article I mentioned above is its characterization of the difference of two order statistics from a $U(0,1)$ distribution: If $U_1, U_2, \ldots U_n \overset{\text{iid}}{\...
dmk's user avatar
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1 vote
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Expectation and variance of Y

You could use the law of total variance, which says that $$\operatorname{Var}[Y]=E[\operatorname{Var}[Y|X]]+\operatorname{Var}[E[Y|X]]$$ As $Y|\{X=x\}\sim U(0,x)$, we have that $$E[Y|X]=\dfrac{X}{2},\...
Julio Puerta's user avatar
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1 vote
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Related to Laplace Transform and Expectation operator

You can obtain equation (2) because equation (1) transforms directly into equation (2). Let me walk you through the transformation: First you need to recognize that the expectation of an exponential ...
James Shakarji's user avatar
1 vote

Finding the joint distribution of two dependent variables $X_1$ and $X_2=(X_1)^2$

There is no joint probability density function for $X_1$ and $X_2$, as they are not jointly continuous; the support of $(X_1,X_2)$ is on one-dimensional segment of $\mathbb{R}^2$ and hence is not two-...
van der Wolf's user avatar
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1 vote

Almost surely convergence of Bernoulli distribution ($\frac{1}{n}$)

It's written that $X_n \to 1$ a.s. It's false. Indeed, $X_n \to 0$ in probability and hence $X_n$ can't converge to $1$ a.s. It looks like you imply that $limsup X_n = 1$ a.s. Let us prove it ...
Sergei Nikolaev's user avatar
1 vote
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Show: $\text{Var}(Y) = \text{Var}(Y - E[Y | X]) + \text{Var}(E[Y | X])$.

You're almost there. It's useful to note by iterated expectation $$\mathbb{E}\left[ Y \mathbb{E}\left[ Y \mid X \right] \right] = \mathbb{E}\left[ \mathbb{E}\left[ Y \mid X \right]^2 \right]$$ so $$ \...
msantama's user avatar
  • 1,096
1 vote
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Expected rank of a random binary matrix with Bernoulli probability p?

Upon reading the paper I linked in the original post a lot closer, I realize that the author does in fact go over the general case where the matrix entries draw from a Bernoulli distribution with ...
user3667125's user avatar
1 vote

Find the Expected value of the discrete random variable $Z=X+Y$

HINT: Note that $$P(Z = n) = \sum_{k = 1}^n P(X = n - k)P(Y = k) = \frac{30}{\pi 6^n}\sum_{k = 1}^n \frac{6^k}{k^2} $$ as for $k < 1$ and $k > n$ the sum above vanishes. Now if you set $$S(x) = \...
Falcon's user avatar
  • 4,054
1 vote

How should independent random variables be distributed to have their product be distributed as a gaussian?

Here is a partial answer inspired by Robert Israel's idea. Actually, computations will turn out to be carried more easily easier with a positive random variable, that is why we will consider the sign ...
Abezhiko's user avatar
  • 8,551
1 vote
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Continuity of random variable and its CDF relation in $R^d$

$F$ is continuous at $(x_1, \ldots, x_n)$ if and only if $F(x_1-h, \ldots, x_n-h) \to F(x_1, \ldots, x_n)$. The limit of the difference $\lim_{h \to 0} [F(x_1,\ldots, X_n) - F(x_1-h, \ldots, x_n-h)]$ ...
angryavian's user avatar
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