MHB Find Probability: P(Z $\leq$ 9/5)

[mathmari]

Hey!

I saw the following exercise and I wanted to solve it,but I got stuck.. (Worried)

Let $X,Y$ be independent random variables,that follow uniform distribution on $[0,1]$, and let the random variable $Z=X+Y$.

The density of $Z$ is:
$$f_{X+Y}(z)=\int_0^z f_X(x)f_Y(z-x)dx$$

How could we calculate the probability $P(Z \leq \frac{9}{5})$? (Wondering)

 

[I like Serena]

Hey!

I saw the following exercise and I wanted to solve it,but I got stuck.. (Worried)

Let $X,Y$ be independent random variables,that follow uniform distribution on $[0,1]$, and let the random variable $Z=X+Y$.

The density of $Z$ is:
$$f_{X+Y}(z)=\int_0^z f_X(x)f_Y(z-x)dx$$

How could we calculate the probability $P(Z \leq \frac{9}{5})$? (Wondering)

Hi! ;)

Let's evaluate your integral:
\begin{aligned}

f_{Z}(z) = f_{X+Y}(z) &= \int_0^z f_X(x)f_Y(z-x)\,dx \\

&= \begin{cases}
\int_0^z 1 \cdot f_Y(z-x)\,dx &\text{if } 0 \le z < 1 \\
\int_0^1 1 \cdot f_Y(z-x)\,dx &\text{if } 1 \le z \le 2 \\
0 & \text{otherwise}
\end{cases} \\

&= \begin{cases}
\int_0^z 1 \cdot 1\,dx &\text{if } 0 \le z < 1 \\
\int_{z-1}^1 1 \cdot 1\,dx &\text{if } 1 \le z \le 2 \\
0 & \text{otherwise}
\end{cases} \\

&= \begin{cases}
z &\text{if } 0 \le z < 1 \\
2-z &\text{if } 1 \le z \le 2 \\
0 & \text{otherwise}
\end{cases} \\

\end{aligned}

Now:
$$P(Z \le \frac 95) = \int_0^{9/5} f_Z(z)\,dz$$
(Thinking)

 

[chisigma]

Hey!

I saw the following exercise and I wanted to solve it,but I got stuck.. (Worried)

Let $X,Y$ be independent random variables,that follow uniform distribution on $[0,1]$, and let the random variable $Z=X+Y$.

The density of $Z$ is:
$$f_{X+Y}(z)=\int_0^z f_X(x)f_Y(z-x)dx$$

How could we calculate the probability $P(Z \leq \frac{9}{5})$? (Wondering)

It is easy to arrive at...

$\displaystyle f_{Z} (x) = \begin{cases} x & \text{if}\ 0 \le x < 1\\ 2 - x & \text{if}\ 1 \le x \le 2\\ 0 & \text{otherwise}\ \end{cases}\ (1)$

... so that is...

$\displaystyle P \{Z < \frac{9}{5} \} = \int_{0}^{1} x\ d x + \int_{1}^{\frac{9}{5}} (2 - x)\ dx = \frac{1}{2} + \frac{24}{50} = \frac{49}{50}\ (2)$

Kind regards

$\chi$ $\sigma$

 

[mathmari]

Hi! ;)

Let's evaluate your integral:
\begin{aligned}

f_{Z}(z) = f_{X+Y}(z) &= \int_0^z f_X(x)f_Y(z-x)\,dx \\

&= \begin{cases}
\int_0^z 1 \cdot f_Y(z-x)\,dx &\text{if } 0 \le z < 1 \\
\int_0^1 1 \cdot f_Y(z-x)\,dx &\text{if } 1 \le z \le 2 \\
0 & \text{otherwise}
\end{cases} \\

&= \begin{cases}
\int_0^z 1 \cdot 1\,dx &\text{if } 0 \le z < 1 \\
\int_{z-1}^1 1 \cdot 1\,dx &\text{if } 1 \le z \le 2 \\
0 & \text{otherwise}
\end{cases} \\

&= \begin{cases}
z &\text{if } 0 \le z < 1 \\
2-z &\text{if } 1 \le z \le 2 \\
0 & \text{otherwise}
\end{cases} \\

\end{aligned}

Now:
$$P(Z \le \frac 95) = \int_0^{9/5} f_Z(z)\,dz$$
(Thinking)

Could you explain why the equality $$\int_0^z f_X(x)f_Y(z-x)\,dx = \begin{cases}
\int_0^z 1 \cdot f_Y(z-x)\,dx &\text{if } 0 \le z < 1 \\
\int_0^1 1 \cdot f_Y(z-x)\,dx &\text{if } 1 \le z \le 2 \\
0 & \text{otherwise}
\end{cases} $$ stands?? Why do we have to take these cases?? (Wondering)

And also, do we replace at the integral $f_X(x)$ with $1$ using the formula $$f(x)=\left\{\begin{matrix}
\frac{1}{b-a} & , a \leq x\leq b\\
0 &, x<a \text{ or } x>b
\end{matrix}\right.$$ ?? (Wondering)

 

[I like Serena]

Could you explain why the equality $$\int_0^z f_X(x)f_Y(z-x)\,dx = \begin{cases}
\int_0^z 1 \cdot f_Y(z-x)\,dx &\text{if } 0 \le z < 1 \\
\int_0^1 1 \cdot f_Y(z-x)\,dx &\text{if } 1 \le z \le 2 \\
0 & \text{otherwise}
\end{cases} $$ stands?? Why do we have to take these cases?? (Wondering)

And also, do we replace at the integral $f_X(x)$ with $1$ using the formula $$f(x)=\left\{\begin{matrix}
\frac{1}{b-a} & , a \leq x\leq b\\
0 &, x<a \text{ or } x>b
\end{matrix}\right.$$ ?? (Wondering)

Your second question is the answer to the first one. (Wasntme)

Since $f_X$ and $f_Y$ are piece wise functions, we need to distinguish the cases where they are $0$ and non-zero.
That's why the integral needs to be split up into separate cases.
And that's why the bounds can be adjusted to only specify the bounds where $f_X$ respectively $f_Y$ are non-zero. The part where they are zero vanishes. (Mmm)

 

[mathmari]

Your second question is the answer to the first one. (Wasntme)

Since $f_X$ and $f_Y$ are piece wise functions, we need to distinguish the cases where they are $0$ and non-zero.
That's why the integral needs to be split up into separate cases.
And that's why the bounds can be adjusted to only specify the bounds where $f_X$ respectively $f_Y$ are non-zero. The part where they are zero vanishes. (Mmm)

Ahaa..

At the formula $$f(x)=\left\{\begin{matrix}
\frac{1}{b-a} & , a \leq x\leq b\\
0 &, x<a \text{ or } x>b
\end{matrix}\right.$$ we take cases as for $x$, and the variable of $f$ is $x$.

But at the formula
$$\int_0^z f_X(x)f_Y(z-x)\,dx = \begin{cases}
\int_0^z 1 \cdot f_Y(z-x)\,dx &\text{if } 0 \le z < 1 \\
\int_0^1 1 \cdot f_Y(z-x)\,dx &\text{if } 1 \le z \le 2 \\
0 & \text{otherwise}
\end{cases} $$ why do we take cases as for $z$ although the variable of $f_X$ is $x$?? (Wondering)

 

[I like Serena]

Ahaa..

At the formula $$f(x)=\left\{\begin{matrix}
\frac{1}{b-a} & , a \leq x\leq b\\
0 &, x<a \text{ or } x>b
\end{matrix}\right.$$ we take cases as for $x$, and the variable of $f$ is $x$.

But at the formula
$$\int_0^z f_X(x)f_Y(z-x)\,dx = \begin{cases}
\int_0^z 1 \cdot f_Y(z-x)\,dx &\text{if } 0 \le z < 1 \\
\int_0^1 1 \cdot f_Y(z-x)\,dx &\text{if } 1 \le z \le 2 \\
0 & \text{otherwise}
\end{cases} $$ why do we take cases as for $z$ although the variable of $f_X$ is $x$?? (Wondering)

The range of $x$ in the integral is $[0,z]$.

If $0 \le z < 1$, then that implies that $0 \le x < 1$. Therefore $f_X(x) = 1$.

If $1 \le z \le 2$, then that implies that $0 \le x \le 2$.
However, we only have $f_X(x) = 1$ if $0 \le x \le 1$.
Therefore we need to limit the upper boundary to $1$ instead of $z$.
Then $f_X(x) = 1$ on the complete range of the integral.
(Wasntme)

 

[mathmari]

The range of $x$ in the integral is $[0,z]$.

If $0 \le z < 1$, then that implies that $0 \le x < 1$. Therefore $f_X(x) = 1$.

If $1 \le z \le 2$, then that implies that $0 \le x \le 2$.
However, we only have $f_X(x) = 1$ if $0 \le x \le 1$.
Therefore we need to limit the upper boundary to $1$ instead of $z$.
Then $f_X(x) = 1$ on the complete range of the integral.
(Wasntme)

Knowing that $$f_X(x) = \begin{cases}
1&\text{if } 0 \le x \le 1 \\
0 & \text{otherwise}
\end{cases}$$
we have the following:

$$f_Z(z)=\int_0^z f_X(x)f_Y(z-x)\,dx = \begin{cases}
\int_0^z f_X(x) \cdot f_Y(z-x)\,dx &\text{if } 0 \le z < 1 \\
\int_0^z f_X(x) \cdot f_Y(z-x)\,dx &\text{if } 1 \le z \le 2 \\
0 & \text{otherwise}
\end{cases} \\ = \begin{cases}
\int_0^z 1 \cdot f_Y(z-x)\,dx &\text{if } 0 \le z < 1 \\
\int_0^1 f_X(x) \cdot f_Y(z-x)\,dx +\int_1^z f_X(x) \cdot f_Y(z-x)\,dx &\text{if } 1 \le z \le 2 \\
0 & \text{otherwise}
\end{cases} \\ = \begin{cases}
\int_0^z 1 \cdot f_Y(z-x)\,dx &\text{if } 0 \le z < 1 \\
\int_0^1 1 \cdot f_Y(z-x)\,dx +\int_1^z 0 \cdot f_Y(z-x)\,dx &\text{if } 1 \le z \le 2 \\
0 & \text{otherwise}
\end{cases} \\ = \begin{cases}
\int_0^z 1 \cdot f_Y(z-x)\,dx &\text{if } 0 \le z < 1 \\
\int_0^1 1 \cdot f_Y(z-x)\,dx &\text{if } 1 \le z \le 2 \\
0 & \text{otherwise}
\end{cases}$$

Then knowing that $$f_Y(x) = \begin{cases}
1&\text{if } 0 \le x \le 1 \\
0 & \text{otherwise}
\end{cases}$$ and
$0 \le x \le 1 \Rightarrow -1 \le x \le 0 \Rightarrow z-1 \le z-x \le z$

we have the following:

$$f_Z(z)=\int_0^z f_X(x)f_Y(z-x)\,dx = \begin{cases}
\int_0^z 1 \cdot f_Y(z-x)\,dx &\text{if } 0 \le z < 1 \\
\int_0^1 1 \cdot f_Y(z-x)\,dx &\text{if } 1 \le z \le 2 \\
0 & \text{otherwise}
\end{cases} \\ = \begin{cases}
\int_0^z 1 \cdot 1\,dx &\text{if } 0 \le z < 1 \\
\int_0^{z-1} 1 \cdot f_Y(z-x)\,dx + \int_{z-1}^1 1 \cdot f_Y(z-x)\,dx &\text{if } 1 \le z \le 2 \\
0 & \text{otherwise}
\end{cases} \\ = \begin{cases}
\int_0^z 1\,dx &\text{if } 0 \le z < 1 \\
\int_0^{z-1} 1 \cdot 0\,dx + \int_{z-1}^1 1 \cdot 1\,dx &\text{if } 1 \le z \le 2 \\
0 & \text{otherwise}
\end{cases} \\ = \begin{cases}
\int_0^z 1\,dx &\text{if } 0 \le z < 1 \\
\int_{z-1}^1 1\,dx &\text{if } 1 \le z \le 2 \\
0 & \text{otherwise}
\end{cases} \Rightarrow \\ f_Z(z)= \begin{cases}
z &\text{if } 0 \le z < 1 \\
2-z\,dx &\text{if } 1 \le z \le 2 \\
0 & \text{otherwise}
\end{cases}$$

Therefore, $$P \left ( Z \leq \frac{9}{5} \right ) =\int_0^{\frac{9}{5}} f_Z(z)dz=\int_0^1 z dz+\int_1^{\frac{9}{5}} (2-z)dz= \left [ \frac{z^2}{2} \right ]_0^1+ \left [ 2z-\frac{z^2}{2} \right ]_1^{\frac{9}{5}}= \\ \frac{1}{2}+2\frac{9}{5}-\frac{1}{2} \left ( \frac{9}{5} \right )^2-2+\frac{1}{2}=\frac{18}{5}-\frac{81}{50}-1=\frac{180-81-50}{50}=\frac{49}{50}$$

Is this correct?? (Wondering)

 

[I like Serena]

Yep! (Nod)

(Except for $0≤x≤1⇒−1≤x≤0$. )[/color]

 

[mathmari]

Yep! (Nod)

Great! (flower) Thank you very much! (Yes)

(Except for $0≤x≤1⇒−1≤x≤0$. )[/color]

I forgot a minus sign.. (Lipssealed)