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Let the value of your car be M dollars. When you get into an accident, the amount of damage to your car is a random variable X with probability density function

\begin{align} f(x)=\begin{cases}2(M-x)/M^2 & 0 < x < M \\0 & x<0 \ or \ x>M\end{cases} \end{align}

You have insurance with an M/2 dollar deductable. That is, you pay only the first M/2 dollars of damage, and the insurance company pays any amount of damage over M/2 dollars. Let Y be the amount you pay for the damage.

(c) Find the probabilities $P(Y<y) \ for \ y<M/2, P(Y=M/2),$ and $P(Y<y) \ for \ y>M/2.$ Based on these, write a formula for and sketch the graph of the cumulative distribution function $F_Y$ of the random variable Y.

I have determined that Y is a mixed random variable with the following cdf: \begin{align} F_Y(y)=\begin{cases}0 & y<0 \\ \frac{2}{M}y -\frac{1}{M^2}y^2 & 0 \leq y < \frac{M}{2} \\1 & y \geq \frac{M}{2}\end{cases} \end{align}

So, $y=M/2$ is the point where cdf has a jump, how do I compute the $P(Y=M/2)$: like for a discrete random variable (the size of a jump) $P(Y=M/2)= 1-P(Y<\frac{M}{2})=1-3/4=1/4$ or like for a continuous random variable P(Y=M/2)=0?

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  • $\begingroup$ Hint: $P(Y = y) = P(Y \leq y) - P(Y < y) = F_Y(y) - F_Y(y-)$. With $h(y-)$ denoting the left limit of the function $h$ in $y$. $\endgroup$ – Furrer Oct 13 '18 at 14:25
  • $\begingroup$ @Furrer $P(Y=\frac{M}{2})=P(Y≤\frac{M}{2})−P(Y<\frac{M}{2})=F_Y(\frac{M}{2})−F_Y(\frac{M}{2}^−)=1 - \lim_{t->\frac{M}{2}^-} = 1 - \frac{3}{4} $so it will be $\frac{1}{4}$? $\endgroup$ – dxdydz Oct 13 '18 at 14:55
  • $\begingroup$ Yes. You can think of it the following way: Y is a continuous random variable except in the point $M/2$. $\endgroup$ – Furrer Oct 13 '18 at 15:02

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