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## Q. 2.10

Find the probability density function for the mixed random variable Y of Examples 2.3 and 2.5, having

$F_{Y} (u) = 0.05 (u + 4) U (u) + (0.8 – 0.05 u) U (u – 10)$

## Verified Solution

Differentiating this cumulative distribution function gives

$p_{Y} (u) = 0.05U (u)+0.05 (u+4) \delta (u) – 0.05 U (u – 10)+(0.8-0.05u) \delta (u-10)$

which can be rewritten as

$p_{Y} (u) = 0.05[U (u)-U (u – 10)]+0.2 \delta (u)+0.3 \delta (u-0)$

Or

$p_{Y} (u) = 0.05U (u) U (10 – u)]+0.2 \delta (u)+0.3 \delta (u-0)$

but these final forms require using some properties of the Dirac delta function. In particular, $0.05(u + 4) \delta (u) = 0.05 \delta (u)$, because the terms are equal at $u = 0$ and both are zero everywhere else. Similarly, $(0.8 -0.05 u) \delta (u – 10) = 0.3 \delta (u – 10)$, because they match at the sole nonzero point of $u = 10$.