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If $X$ and $Y$ are i.i.d. exponential as $f_X(x)=\lambda e^{-\lambda x}$ and $f_Y(y)=\lambda e^{-\lambda y}$, and $Z=\min(X,Y)$, then we may say that probability of event $Z=X$ or $Z=Y$ is $\frac 1 2$. This may be valid for any kind of distribution.

If $X$ and $Y$ are independent but not identical exponential as $f_X(x)=\lambda_x e^{-\lambda_x x}$ and $f_Y(y)=\lambda_y e^{-\lambda_y y}$, and $Z=\min(X,Y)$, then how I can find the probability of event $Z=X$ or $Z=Y$ ?.

Most importantly, I need how to find this probability when $X$ and $Y$ have any kind of continuous distributions which are independent.

Jean Marie
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Frey
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1 Answers1

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I "think" I understand you are asking, and the probability that $X$ occurs before $Y$ is

$$P(X < Y)=\frac{\lambda_x}{\lambda_x+\lambda_y}$$

and the probability that $Y$ occurs before $X$ is

$$P(Y < X)=\frac{\lambda_y}{\lambda_x+\lambda_y}$$

For the exponential distribution, you can compute this probability like this:

$$ \mathbb{Pr}(X\ will\ finish\ before\ Y) = \int_0^{\infty}\lambda_x e^{-\lambda_x t}e^{-\lambda_y t} dt = \int_0^{\infty}e^{-(\lambda_x + \lambda_y)t} dt = \frac{\lambda_x}{\lambda_x + \lambda_y}$$

And, in general

$$\mathbb{Pr}(X\ will\ finish\ before\ Y) = \int_0^{\infty}b_x(t)R_y(t) dt,$$

where $b_x(t)$ is the density of $X$ and $R_y(t)$ is the reliability function of $Y$ -- i.e., $R_y(t)$ = 1 - $F_y(t)$.

PiE
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  • Can you please give me some steps how you derive it? Because I will not always deal with exponential. better to know some general way if you can provide. – Frey May 17 '17 at 02:55
  • For the exponential distribution: $\mathbb{Pr}(X\ will\ finish\ before\ Y) = \int_0^{\infty}\lambda_x e^{-\lambda_x t}e^{-\lambda_y t} dt = \int_0^{\infty}e^{-(\lambda_x + \lambda_y)t} dt = \frac{\lambda_x}{\lambda_x + \lambda_y}$ – PiE May 17 '17 at 03:02
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    I like your general expression, thanks @PIE – Frey May 17 '17 at 03:18