document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1164362>Question 1164362</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Suppose X and Y are independent random variables. X is exponentially distributed with mean 1/a) and Y is exponentially distributed with mean 1/b. For a random variable with mean µ define the rate of the random variable as 1/µ. a) Find the pdf of the random variable Z = min(X,Y).  b) What is the mean of this distribution? c) What is the relationship between the rate of min(X,Y) and the rates of X and Y.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #854354 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(2261)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=854354\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> To solve for the distribution of the minimum of two independent exponential random variables, we first identify their parameters. \r<BR>\n" );
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document.write( "For an exponential distribution, the **rate** $\lambda$ is the reciprocal of the **mean** $\mu$. <BR>\n" );
document.write( "* $X$ has mean $1/a$, so its rate $\lambda_X = a$.<BR>\n" );
document.write( "* $Y$ has mean $1/b$, so its rate $\lambda_Y = b$.\r<BR>\n" );
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document.write( "### **a) Find the PDF of $Z = \min(X, Y)$**\r<BR>\n" );
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document.write( "To find the probability density function (PDF), it is easiest to start with the **Survival Function** $P(Z > z)$. <BR>\n" );
document.write( "The minimum of two values is greater than $z$ if and only if **both** values are greater than $z$:<BR>\n" );
document.write( "$$P(Z > z) = P(\min(X, Y) > z) = P(X > z \text{ and } Y > z)$$\r<BR>\n" );
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document.write( "Since $X$ and $Y$ are independent:<BR>\n" );
document.write( "$$P(Z > z) = P(X > z) \cdot P(Y > z)$$<BR>\n" );
document.write( "$$P(Z > z) = e^{-az} \cdot e^{-bz} = e^{-(a+b)z}$$\r<BR>\n" );
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document.write( "The Cumulative Distribution Function (CDF) is $F_Z(z) = 1 - P(Z > z) = 1 - e^{-(a+b)z}$. To find the PDF, we take the derivative with respect to $z$:<BR>\n" );
document.write( "$$f_Z(z) = \frac{d}{dz} \left( 1 - e^{-(a+b)z} \right) = (a+b)e^{-(a+b)z} \quad \text{for } z \ge 0$$\r<BR>\n" );
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document.write( "### **b) What is the mean of this distribution?**\r<BR>\n" );
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document.write( "The resulting PDF is clearly that of another exponential distribution with a new rate parameter $\lambda_Z = a + b$. The mean of an exponential distribution is the reciprocal of its rate:<BR>\n" );
document.write( "$$E[Z] = \frac{1}{a + b}$$\r<BR>\n" );
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document.write( "### **c) Relationship between the rates**\r<BR>\n" );
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document.write( "The **rate** of the minimum of two independent exponential random variables is the **sum** of the individual rates:<BR>\n" );
document.write( "$$\text{Rate}(Z) = \text{Rate}(X) + \text{Rate}(Y)$$\r<BR>\n" );
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document.write( "**Intuition:** Imagine two independent lightbulbs. If the time it takes for bulb $A$ to burn out has rate $a$ and bulb $B$ has rate $b$, the \"system\" (which fails as soon as the first bulb goes out) fails at a combined, faster rate of $a + b$. </SPAN>\n" );
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