# Exam P Practice Problem 99 – When Random Loss is Doubled

Problem 99-A

A business owner faces a risk whose economic loss amount $X$ follows a uniform distribution over the interval $0. In the next year, the loss amount is expected to be doubled and is expected to be modeled by the random variable $Y=2X$.

Suppose that the business owner purchases an insurance policy effective at the beginning of next year with the provision that any loss amount less than or equal to 0.5 is the responsibility of the business owner and any loss amount that is greater than 0.5 is paid by the insurer in full. When a loss occurs next year, determine the expected payment made by the insurer to the business owner.

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \frac{8}{16}$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \frac{9}{16}$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \frac{13}{16}$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \frac{15}{16}$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \frac{17}{16}$

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Problem 99-B

A business owner faces a risk whose economic loss amount $X$ has the following density function:

$\displaystyle f(x)=\frac{x}{2} \ \ \ \ \ \ 0

In the next year, the loss amount is expected to be doubled and is expected to be modeled by the random variable $Y=2X$.

Suppose that the business owner purchases an insurance policy effective at the beginning of next year with the provision that any loss amount less than or equal to 1 is the responsibility of the business owner and any loss amount that is greater than 1 is paid by the insurer in full. When a loss occurs next year, what is the expected payment made by the insurer to the business owner?

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ 0.6667$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ 1.5833$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ 1.6875$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ 1.7500$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ 2.6250$

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# Exam P Practice Problem 84 – When Random Loss is Doubled

Problem 84-A

A business owner faces a risk whose economic loss amount $X$ follows a uniform distribution over the interval $0. In the next year, the loss amount is expected to be doubled and is expected to be modeled by the random variable $Y=2X$.

Suppose that the business owner purchases an insurance policy effective at the beginning of next year with the provision that any loss amount less than or equal to 0.5 is the responsibility of the business owner and any loss amount in excess of 0.5 is the responsibility of the insurer. When a loss occurs next year, what is the expected payment made by the insurer to the business owner?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \frac{4}{16}$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \frac{6}{16}$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \frac{7}{16}$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \frac{8}{16}$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \frac{9}{16}$

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Problem 84-B

A business owner faces a risk whose economic loss amount $X$ has the following density function:

$\displaystyle f(x)=\frac{x}{2} \ \ \ \ \ \ 0

In the next year, the loss amount is expected to be doubled and is expected to be modeled by the random variable $Y=2X$.

Suppose that the business owner purchases an insurance policy effective at the beginning of next year with the provision that any loss amount less than or equal to 1 is the responsibility of the business owner and any loss amount in excess of 1 is the responsibility of the insurer. When a loss occurs next year, what is the expected payment made by the insurer to the business owner?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \frac{5}{12}$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \frac{2}{3}$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \frac{19}{12}$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \frac{27}{16}$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \frac{21}{12}$

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$\copyright \ 2014 \ \ \text{ Dan Ma}$

# Exam P Practice Problem 75 – Travel Time to Work By Train

Both Problem 75-A and Problem 75-B use the following information.

A worker travels to work by train 5 days a week (Monday to Friday). The length of a train ride (in minutes) to work follows a continuous uniform distribution from 10 to 40.

The lengths of the train ride across the days of the week are independent.

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Problem 75-A

What is the probability that the shortest train ride during a work week is between 15 and 20 minutes?

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.025$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.039$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.045$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.053$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.064$

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Problem 75-B

What is the expected value of the longest train ride during a work week?

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 25.0$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 25.9$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 28.2$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 33.3$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 35.7$

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$\copyright \ 2013-2016 \ \ \text{Dan Ma}$

# Exam P Practice Problem 74 – Review of Auto Insurance Claims

Both Problem 74-A and Problem 74-B use the following information.

An insurer issued policies to cover a large number of automobiles. Claim amounts (in thousands) from these policies are independent and are modeled by a continuous uniform distribution on (0,10).

The insurer randomly selects five claims for review.

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Problem 74-A

What is the probability that the minimum claim amount is between 2 thousands and 6 thousands?

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.31$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.32$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.33$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.34$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.75$

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Problem 74-B

What is the expected value of the maximum claim amount?

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5.0$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5.5$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 7.6$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 8.3$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 8.5$

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$\copyright \ 2013 \ \ \text{Dan Ma}$

# Exam P Practice Problem 46 – Finding Moment of a Sum

Problem 46-A

Suppose that $X$ and $Y$ are random losses that are jointly distributed with the following density function:

$\displaystyle f(x,y)=\frac{1}{40} \ (4-x) \ \ \ \ \ \ 0

Find the second moment of the sum of the two losses.

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Problem 46-B

Suppose that $X$ and $Y$ are random losses that are jointly distributed with the following density function:

$\displaystyle f(x,y)=16 \ x^2 \ y \ e^{-2 \ (x \ + \ y)} \ \ \ \ \ \ 0

Find the second moment of the sum of the two losses.

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# Exam P Practice Problem 26 – Uniform Distribution

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This post has no alternate problem. It has one problem with 3 parts.

Problem 26

For an office worker, the length of time, $X$, of the bus ride from home to office follows a uniform distribution from $0$ to $20$ minutes and the length of time, $Y$, of the bus ride from office back to home follows a uniform distribution from $0$ to $25$ minutes.

Suppose that length of bus rides in one direction is independent of the length of bus rides in the other direction.

In recent weeks, the worker finds that the total daily time spent on bus rides always exceeds $20$ minutes.

1. What is the probability that the bus ride to return home will take more than 10 minutes?
2. What is the expected length of the bus rides from office to home?
3. What is the variance of the length of the bus rides from office to home?

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$\copyright \ 2013$

# Exam P Practice Problem 25 – Uniform Distribution

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This post has no alternate problem. It has one problem with 3 parts.

Problem 25

A real estate property owner is at risk for losses due to two different perils affecting her property. There are no other potential losses in addition to these two different types of losses. Let $X$ be the total amount of the losses in a year due to one peril and let $Y$ be the total amount of the losses in a year due to the other peril. Suppose that $X$ and $Y$ are independent and are identically and uniformly distributed from $0$ to $10$.

In recent years, the owner finds that the total annual losses (due to both perils) always exceed 10.

1. What is the probability that the loss $X$ exceeds $5$?
2. What is the expected annual loss $X$?
3. What is the variance of the annual loss $X$?

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# Examples of convolution (discrete case)

Consider the following problems.

Problem 1
Roll a fair die two times. What is the probability that the sum of the two rolls is 5?

Problem 2
There are two independent multiple choice quizzes where each quiz has 5 questions. Each question on the first quiz has 4 choices and each question on the second quiz has 5 choices. Suppose a student answers the questions in the quizzes by pure guessing. What is the probability that the student obtains 5 correct answers in these two quizzes?

Problem 3
For the two independent quizzes in Problem 2, find the probability function for the total number of correct answers. What is the probability that the student get 6 or more correct answers in the two quizzes?

Problem 4
There are two independent multiple choice quizzes where each quiz has 5 questions. There are 5 choices for all questions in these two quizzes. Suppose the student answers the questions in the quizzes by pure guessing. Find the probability function for the total number of correct answers.

All of the above problems are about the independent sum of discrete random variables. We demonstrate the convolution technique using Problem 2.

The Convolution Formula (Discrete Case)
Let $X$ and $Y$ be independent discrete random variables with probability functions $p_X(x)=P(X=x)$ and $p_Y(y)=P(Y=y)$, respectively. Then the following is the probability function of $Z=X+Y$.

\displaystyle \begin{aligned}p_Z(z)&=\sum \limits_{x+y=z} p_X(x) \ p_Y(y) \\&\text{ } \\&=\sum \limits_{x} p_X(x) \ p_Y(z-x) \\&\text{ } \\&=\sum \limits_{y} p_X(z-y) \ p_Y(y) \end{aligned}

Note that the joint probability function of $X$ and $Y$ is $p_{X,Y}(x,y)=p_X(x) \ p_Y(y)$. The convolution formula says that the probability function of the independent sum $Z=X+Y$ is obtained by summing the joint probabiity over the line $x+y=z$.

Problem 2
Let $X$ be the number of correct guesses in quiz 1 and let $Y$ be the number of correct guesses in quiz 2. The variable $X$ has a binomial distribution with parameters $n=5$ and $p=\frac{1}{4}$. The variable $Y$ has a binomial distribution with parameters $n=5$ and $p=\frac{1}{5}$. Let $Z=X+Y$.

\displaystyle \begin{aligned}p_Z(5)&=P(X=0)P(Y=5)+P(X=1)P(Y=4)+P(X=2)P(Y=3) \\&\text{ } \\&\ \ \ +P(X=3)P(Y=2)+P(X=4)P(Y=1)+P(X=5)P(Y=0) \\&\text{ } \\&=\binom{5}{0} \biggl(\frac{1}{4}\biggr)^0 \biggl(\frac{3}{4}\biggr)^5 \ \ \binom{5}{5} \biggl(\frac{1}{5}\biggr)^5 \biggl(\frac{4}{5}\biggr)^0 \\&\ \ \ +\binom{5}{1} \biggl(\frac{1}{4}\biggr)^1 \biggl(\frac{3}{4}\biggr)^4 \ \ \binom{5}{4} \biggl(\frac{1}{5}\biggr)^4 \biggl(\frac{4}{5}\biggr)^1 \\&\ \ \ +\binom{5}{2} \biggl(\frac{1}{4}\biggr)^2 \biggl(\frac{3}{4}\biggr)^3 \ \ \binom{5}{3} \biggl(\frac{1}{5}\biggr)^3 \biggl(\frac{4}{5}\biggr)^2 \\&\ \ \ +\binom{5}{3} \biggl(\frac{1}{4}\biggr)^3 \biggl(\frac{3}{4}\biggr)^2 \ \ \binom{5}{2} \biggl(\frac{1}{5}\biggr)^2 \biggl(\frac{4}{5}\biggr)^3 \\&\ \ \ +\binom{5}{4} \biggl(\frac{1}{4}\biggr)^4 \biggl(\frac{3}{4}\biggr)^1 \ \ \binom{5}{1} \biggl(\frac{1}{5}\biggr)^1 \biggl(\frac{4}{5}\biggr)^4 \\&\ \ \ +\binom{5}{5} \biggl(\frac{1}{4}\biggr)^5 \biggl(\frac{3}{4}\biggr)^0 \ \ \binom{5}{0} \biggl(\frac{1}{5}\biggr)^0 \biggl(\frac{4}{5}\biggr)^5 \\&\text{ } \\&=\frac{129367}{3200000}=0.0404271875 \end{aligned}

Problem 3
Let $Z=X+Y$ where $X$ and $Y$ are as in Problem 2. The following probabilities are obtained by applying the convolution formula.

$\displaystyle P(Z=0)=\frac{248832}{3200000}=0.07776$

$\displaystyle P(Z=1)=\frac{725760}{3200000}=0.2268$

$\displaystyle P(Z=2)=\frac{950400}{3200000}=0.297$

$\displaystyle P(Z=3)=\frac{735840}{3200000}=0.22995$

$\displaystyle P(Z=4)=\frac{373020}{3200000}=0.11656875$

$\displaystyle P(Z=5)=\frac{129367}{3200000}=0.040427$

$\displaystyle P(Z=6)=\frac{31085}{3200000}=0.009714$

$\displaystyle P(Z=7)=\frac{5110}{3200000}=0.001596875$

$\displaystyle P(Z=8)=\frac{550}{3200000}=0.000171875$

$\displaystyle P(Z=9)=\frac{35}{3200000}=0.0000109375$

$\displaystyle P(Z=10)=\frac{1}{3200000}$

$\displaystyle P(Z \ge 6)=\frac{36781}{3200000}=0.011494$

Problem 4
Let $Z=X+Y$ where both $X$ and $Y$ are binomial with parameter $n=5$ and $p=\frac{1}{5}$. The independent sum $Z$ is binomial with $n=10$ and $p=\frac{1}{5}$. The following probabilities are also exercises for using the convolution formula.

$\displaystyle P(Z=0)=\frac{1048576}{9765625}=0.10737$

$\displaystyle P(Z=1)=\frac{2621440}{9765625}=0.26844$

$\displaystyle P(Z=2)=\frac{2949120}{9765625}=0.30199$

$\displaystyle P(Z=3)=\frac{1966080}{9765625}=0.20133$

$\displaystyle P(Z=4)=\frac{860160}{9765625}=0.08808$

$\displaystyle P(Z=5)=\frac{258048}{9765625}=0.02642$

$\displaystyle P(Z=6)=\frac{53760}{9765625}=0.00551$

$\displaystyle P(Z=7)=\frac{7680}{9765625}=0.000786$

$\displaystyle P(Z=8)=\frac{720}{9765625}=0.0000737$

$\displaystyle P(Z=9)=\frac{40}{9765625}$

$\displaystyle P(Z=10)=\frac{1}{9765625}$

# Examples of convolution (continuous case)

The method of convolution is a great technique for finding the probability density function (pdf) of the sum of two independent random variables. We state the convolution formula in the continuous case as well as discussing the thought process. Some examples are provided to demonstrate the technique and are followed by an exercise.

The Convolution Formula (Continuous Case)
Let $X$ and $Y$ be independent continuous random variables with pdfs $f_X(x)$ and $f_Y(y)$, respectively. Let $Z=X+Y$. Then the following is the pdf of $Z$.

\displaystyle \begin{aligned}f_Z(z)&=\int_{-\infty}^\infty f_Y(z-x) f_X(x) \ dx \\&=\int_{-\infty}^\infty f_X(z-y) f_Y(y) \ dy \end{aligned}

Let’s look at the thought process behind the formula. Since $X$ and $Y$ are independent, the joint pdf of $X$ and $Y$ is $f_{X,Y}(x,y)=f_X(x) \ f_Y(y)$. The pdf of $Z$ is simply the sum of the “joint density” at the points of the line $z=x+y$. In Figure 1 below, every point at the line $z=x+y$ is of the form $(x,z-x)$. The joint density at each such point is $f_X(x) \ f_Y(z-x)$. Summing the values of these joint density produces the probability density function of $Z$.

Setting the limits of the integral depends on knowing the range of possible values of $x$ or $y$ for a given line $z=x+y$. If $X$ and $Y$ can only take on positive values, then for a given line $z=x+y$, both $x$ or $y$ can range from $0$ to $z$ (see Example 1 below).

Example 1
Let $X$ and $Y$ be independent exponentially distributed variables with common density $f(x)=\alpha e^{-\alpha x}$ where $\alpha>0$. Then the following is the pdf of $Z=X+Y$.

\displaystyle \begin{aligned}f_Z(z)&=\int_{0}^z f_Y(z-x) f_X(x) \ dx \\&=\int_{0}^z \alpha e^{-\alpha (z-x)} \alpha e^{-\alpha x} \ dx \\&=\int_{0}^z \alpha^2 e^{-\alpha z} \ dx =\alpha^2 \ z \ e^{-\alpha z} \end{aligned}

The above pdf indicates that the independent sum of two identically distributed exponential variables has a Gamma distribution with parameters $2$ and $\alpha$.

Example 2
Let $X$ and $Y$ be independent uniformly distributed variables, $U(0,10)$ and $U(0,20)$, respectively. The pdf of $Z=X+Y$ is:

$\displaystyle f_Z(z)=\left\{\begin{matrix} \displaystyle \frac{z}{200}&\ \ \ \ \ \ 0 \le z <10 \\{\text{ }}& \\{\displaystyle \frac{1}{20}}&\ \ \ \ \ \ 10 \le z <20 \\{\text{ }}& \\{\displaystyle \frac{30-z}{200}}&\ \ \ \ \ \ 20 \le z <30 \end{matrix}\right.$

The convolution formula is applied three times. For the first case, the line $x+y=z$ ranges in $0 \le z < 10$. For each such line, we have $0. Figure 2 below is a representative diagram.

\displaystyle \begin{aligned}f_Z(z)&=\int_{0}^{z} f_Y(z-x) f_X(x) \ dx \\&=\int_{0}^{z} \frac{1}{20} \frac{1}{10} \ dx =\frac{z}{200} \end{aligned}

For the second case, the line $x+y=z$ ranges in $10 \le z < 20$. For each such line, $x$ ranges from $0$ to $10$. Figure 3 below is a representative diagram.

\displaystyle \begin{aligned}f_Z(z)&=\int_{0}^{10} f_Y(z-x) f_X(x) \ dx \\&=\int_{0}^{10} \frac{1}{10} \frac{1}{20} \ dx =\frac{1}{20} \end{aligned}

For the third case, the line $x+y=z$ ranges in $20 \le z < 30$. Figure 4 below is a representative diagram.

\displaystyle \begin{aligned}f_Z(z)&=\int_{z-20}^{10} f_Y(z-x) f_X(x) \ dx \\&=\int_{z-20}^{10} \frac{1}{10} \frac{1}{20} \ dx =\frac{30-z}{200} \end{aligned}

The following is the graph of the pdf of $Z=X+Y$.

Exercise
Suppose that $X$ is an exponentially distributed variable with pdf $f(x)=\alpha e^{-\alpha x}$ and $Y$ has the uniform distribution $U(0,1)$. Find the pdf of the independent sum $Z=X+Y$.

$\displaystyle f_Z(z)=\left\{\begin{matrix} \displaystyle 1-e^{-\alpha z}&\ \ \ \ \ \ 0 \le z <1 \\{\text{ }}& \\{\text{ }}& \\{\displaystyle e^{-\alpha z}(e^\alpha-1)}&\ \ \ \ \ \ 1 \le z <\infty \end{matrix}\right.$

# Independent sum of uniform

Consider the folowing problems.

• Problem A. Suppose that the losses under two insurance policies are independent such that the loss from each policy follows a continuous uniform distribution on the interval from 0 to 2. Find the probability density function and the cumulative distribution function of the sum of the losses.
• Problem B. Suppose that the losses under two insurance policies are independent such that the loss from one policy follows a continuous uniform distribution on the interval from 0 to 10 and the loss from the other policy is continuously uniformly distributed on the interval from 0 to 20. Find the probability density function and the cumulative distribution function of the sum of the losses.

Discussion of Problem A
Finding information about the probability distribution of the sum of several independent random variables is a basic topic in probability theory and is an important one in the SOA Exam P. One elementary method in finding the probability distribution of an independent sum is through the moment generating functions (e.g. independent sum of normal distributions). Another technique is convolution. For the independent sum of two uniform distributions, we can actually obtain the distribution by a graphical approach. The method of convolution can be applied to the examples discussed here.

Let $X$ and $Y$ be independent such that each follows $U(0,2)$. Let $Z=X+Y$. Then the joint pdf of $X$ and $Y$ is $\displaystyle f(x,y)=\frac{1}{4}$ over $0 \le x \le 2$ and $0 \le y \le 2$. The following is the cumulative distribution function of $Z=X+Y$:

$\displaystyle F_Z(z)=\left\{\begin{matrix}0& \ \ \ \ \ \ z<0 \\{\text{ }}& \\{\displaystyle \frac{z^2}{8}}&\ \ \ \ \ \ 0 \le z <2 \\{\text{ }}& \\{\displaystyle 1-\frac{(4-z)^2}{8}}&\ \ \ \ \ \ 2 \le z <4 \\{\text{ }}& \\{1}&\ \ \ \ \ \ 4 \le z \end{matrix}\right.$

For the case $0 \le z <2$, $F_Z(z)$ is obtained by multiplying the shaded area in the following diagram (Figure 1) by the joint pdf $f(x,y)=\frac{1}{4}$. Note that the area of the shaded region in Figure 1 is $\displaystyle \frac{z^2}{2}$.

For the case $2 \le z <4$, $F_Z(z)$ is obtained by multiplying the complement of the shaded area in the following diagram (Figure 2) by the joint pdf $f(x,y)=\frac{1}{4}$. Note that the area of the shaded region in Figure 2 is $\displaystyle \frac{(4-z)^2}{2}$. The complement of the shaded region is $\displaystyle 4-\frac{(4-z)^2}{2}$.

The following is the pdf of $Z=X+Y$.

$\displaystyle f_Z(z)=\left\{\begin{matrix}\displaystyle \frac{z}{4}&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \le z <2 \\{\text{ }}& \\{\displaystyle \frac{4-z}{4}}&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2 \le z <4 \end{matrix}\right.$

Suppose $X$ and $Y$ are $U(0,10)$ and $U(0,20)$, respectively. Suppose that $Z=X+Y$ is an independent sum. Then the cdf and pdf of $Z$ are:
$\displaystyle F_Z(z)=\left\{\begin{matrix}0& \ \ \ \ \ \ z<0 \\{\text{ }}& \\{\displaystyle \frac{z^2}{400}}&\ \ \ \ \ \ 0 \le z <10 \\{\text{ }}& \\{\displaystyle \frac{z}{20}-\frac{1}{4}}&\ \ \ \ \ \ 10 \le z <20 \\{\text{ }}& \\{\displaystyle 1-\frac{(30-z)^2}{400}}&\ \ \ \ \ \ 20 \le z <30 \\{\text{ }}& \\{1}&\ \ \ \ \ \ 30 \le z \end{matrix}\right.$
$\displaystyle f_Z(z)=\left\{\begin{matrix} \displaystyle \frac{z}{200}&\ \ \ \ \ \ 0 \le z <10 \\{\text{ }}& \\{\displaystyle \frac{1}{20}}&\ \ \ \ \ \ 10 \le z <20 \\{\text{ }}& \\{\displaystyle \frac{30-z}{200}}&\ \ \ \ \ \ 20 \le z <30 \end{matrix}\right.$