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Interpretation of a PDF

You need the pdf to integrate to $1$ over its support, which here is $x \ge 0$. Since $\int\limits_0^\infty e^{-2x}\, dx = \frac12$ and $\frac{1}{\frac12}=2$, the corresponding pdf is $2e^{-2x}$ $\...
Henry's user avatar
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Laplace transform of probability density functions

In your counter example $F(-2)$ is a diverging integral. Actually if $\mu$ is a probability on $R^+$ there exists a maximal $A\in[0,\infty]$ such that the Laplace transform is real analytic on $W=(-A,+...
Letac Gérard's user avatar
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If $\text P[s+t<τ]=\text P[s<τ]\text P[t<τ]$ and $\mathcal F_t:=σ(\bigcup_{s\le t}\{τ\le s\})$, can we factorize $\text P[s+t<τ\mid\mathcal F_s]$?

Unfortunately, Michael's answer is not really the expression what I was looking for. However, we should have $$\left.\operatorname P[s+t<\tau\mid\mathcal F_s]\right|_{\{\;s\;<\;\tau\;\}}=\left.\...
0xbadf00d's user avatar
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If $\text P[s+t<τ]=\text P[s<τ]\text P[t<τ]$ and $\mathcal F_t:=σ(\bigcup_{s\le t}\{τ\le s\})$, can we factorize $\text P[s+t<τ\mid\mathcal F_s]$?

Generally, for any nonnegative random variable $\tau:\Omega\rightarrow[0,\infty)$ and any $s\geq 0$ we have $$\mathcal{F}_s=\sigma(\cup_{v \in [0,s]}\sigma(\{\tau \leq v\})) = \sigma(Y)$$ where we ...
Michael's user avatar
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Excess waiting time given two exponential variables

I think your calculation is correct. Below is an alternate approach that leverages the memorylessness property of the exponential distribution to avoid integrating by parts. 30 minutes would be the ...
angryavian's user avatar
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Joint Central Moments of Bivariate Exponential Distribution

Difficult to answer to the question 'which question should I raise?' If $(X,Y)$ is a pair of dependent standard exponential random variables, then the law of $(U,V)=(e^{-X},e^{-Y})$ is a copula, and ...
Letac Gérard's user avatar
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Solving a stochastic equation by characteristic functions

I got it. We have to use the Mellin-transform mentioned in the Wikipedia article about the distribution of the product of two random variables. The Wikipedia-article quotes the property, that for ...
user1047209's user avatar
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Independence of maximums of independent random iid exponential variables.

however I get WLOG $m>n$: $$ \mathbb{P}[M_n \leq t, M_m \leq t] = \mathbb{P}[X_1,...,X_m \leq t] = F_{X_1}(t)^m \neq F_{X_1}(t)^{n+m} = \mathbb{P}[M_n \leq t] \mathbb{P}[M_m \leq t]$$ Yes, but ...
Graham Kemp's user avatar
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Independence of maximums of independent random iid exponential variables.

Another, purely combinatorial approach: suppose $n>m$ and you order your variables in increasing order and mark them $1,2,\dots, n$. There are $n!$ ways to arrange them, but only $(n-1)!$ ways such ...
van der Wolf's user avatar
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Independence of maximums of independent random iid exponential variables.

Assume $0<m<n$. By symmetry, we have $\mathbb{P}(X_m>X_1,\dots,X_{m-1})=\frac1m$ and $\mathbb{P}(X_n>X_1,\dots,X_{n-1})=\frac1n$ . On the other hand, we have for $0<x<y$ $$(*){:={\...
van der Wolf's user avatar
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