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Let $X$ be exponentially distributed with mean $3$. Then how do we compute the density function of $max\{X,2\}$?

Matt
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    This RV takes a value of 2 with probability $P(X\leq 2)$. The density function for values greater than 2 looks the same as for the exponentially distributed variable. –  Jun 19 '14 at 05:16
  • Thanks! I understand the first argument. Could you explain how the second statement hold? – Matt Jun 19 '14 at 05:19
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    For values greater than two, the density function the same as the density function for $X$ because $\text{max}{X,2}=X$. –  Jun 19 '14 at 05:27

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Let $Z = \max(X,2)$. We calculate $P(Z > c)$. If $c < 2$, this is clearly $1$. If $c \geq 2$, then $Z=X$ so this is just $P(X > c)$.

This specifies the CDF of $Z$ (by $P(Z \leq c) = 1- P(Z>c)$). Now, differentiate it, and put a point mass at the jump at $2$ based on the height of the jump in the CDF.

Batman
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