Given a continuous random variable $X$ and the pdf find....

Question

There is a pattern excercise:

First pattern: We have $X$ and the probability density function is usually:

$$ f(x) = \begin{cases} {ax+bx^2} & {0\lt x \lt 1} \\ 0 & \text{else} \end{cases} $$

Then we are given $E|X| = 0.6$

And usually asks to find:

a) $P\{X>0.5\}$

b) $Var(X)$

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Second pattern:

$$ f(x) = \begin{cases} {cx^4} & {0\lt x \lt 2} \\ 0 & \text{else} \end{cases} $$

Where c is a constant. Find:

a) Mean value of $X$

b) Variance of $X$

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Third pattern (it's more like a reverse problem):

We are given $Y= \sqrt{X}$, find it's cumulative distribution function (CDF) and it's probability density function (pdf).

Update:

The third one is transferred to a new seperate question.

This is not homework. I'm self-studying and in need of some help. Thank you.

Answer 1

I'll start with a step-by-step explanation for the first two, as you say those are more important. Later I'll add the third if I'm able.

The key to solving both of the first two problems is to remember that the pdf for every probability distribution must sum/integrate to one.

First problem:

We are given: $$ f(x) = \begin{cases} {ax+bx^2} & {0\lt x \lt 1} \\ 0 & \text{else} \end{cases} $$

So, we know that: $$ \int_0^1 ax+bx^2\; dx = 1 $$ So step one becomes: $$ \int_0^1 ax+bx^2\; dx = 1\\ \frac{ax^2}{2}+\frac{bx^3}{3}\bigg|_0^1 = 1\\ \frac{a}{2} + \frac{b}{3} = 1\\ 3a+2b=6 $$ We also know that $E(X) = 0.6$ As the definition of the expectation is: $E(X) = \int_{-\infty}^\infty xf(x)\;dx$ we know: $$ \int_0^1 ax^2+bx^3\; dx = 0.6\\ \frac{ax^3}{3}+\frac{bx^4}{4}\bigg|_0^1 = 0.6\\ \frac{a}{3} + \frac{b}{4} = 0.6\\ 4a+3b=7.2 $$ Now we have two equations and two unknowns: $$ 3a+2b=6\\ 4a+3b=7.2\\ a -\frac{3}{4}b = 1.8\\ a=1.8-\frac{3}{4}b\\ 3\left(1.8-\frac{3}{4}b\right) + 2b = 6\\ 5.4 - \frac{9}{4}b + \frac{8}{4}b = 6\\ 5.4 - 6 = -\frac{b}{4}\\ \mathbf{b = -2.4}\\ a=1.8-\frac{3}{4}-2.4\\ \mathbf{a = 3.6} $$ Now that we know the values, the first question is simply $\int_{0.5}^1 3.6x - 2.4x^2\; dx$ which is: $$ \int_{0.5}^1 3.6x-2.4x^2\; dx\\ \frac{3.6x^2}{2}-\frac{2.4x^3}{3}\bigg|_{0.5}^1\\ \left(\frac{3.6\cdot1^2}{2}-\frac{2.4\cdot1^3}{3}\right)-\left(\frac{3.6\cdot{0.5}^2}{2}-\frac{2.4\cdot{0.5}^3}{3}\right)\\ 1 - 0.35 = \mathbf{0.65} $$ The second question asks for the variance. We can use the definition of the variance of $Var(X) = E(X^2) - E(X)^2$. To find the first term we need solve $\int_0^1 ax^3+bx^4\; dx$ which is (following the same derivation as above) $\frac{3.6\cdot1^4}{4}-\frac{2.4\cdot1^5}{5} = \frac{3.6}{4} - \frac{2.4}{5} = 0.42$ So the variance is $0.42-0.6^2 = \mathbf{0.06}$.

Second problem:

You can solve this one the same way as the first. We must have $\int_0^2 cx^4\;dx = 1$. So: $$ \int_0^2 cx^4\;dx = 1\\ \frac{cx^5}{5}\bigg|_0^2 = 1\\ \frac{c\cdot32}{5} = 1\\ \mathbf{c=\frac{5}{32}} $$ The mean value, E(X) is the value of $\int_0^2 \frac{5}{32}x^5\;dx$ which is $\frac{5\cdot2^6}{32\cdot6} = \frac{320}{192} = \mathbf{\frac{5}{3}}$

The variance uses the same relationship as above. $E(X^2) = \int_0^2 \frac{5}{32}x^6\;dx = \frac{5\cdot2^7}{32\cdot7} = \frac{640}{224}$ So the variance is $\frac{640}{224} - {\frac{5}{3}}^2 = \mathbf{\frac{5}{63}}$

Answer 2

The first two pattern are solved by first calculating the missing constants. To do so, use the fact that $\int_{\Omega_x}f(x) = 1$. On the first pattern, you have two constants, so you need an extra equation, which comes from the fact that $EX=0.6$ First Pattern: Solve for $ \int_{\Omega_x}f(x) = 0.5a+b/3 =1 EX=a/3+b/4=0.6 $ Second Pattern: Solve for $\int_{\Omega_x}f(x) = 32c/5 =1$ after calculating $a,b$ you can find the CDF and the required moments. For the third pattern, notice that $P(Y<t)=P(\sqrt{X}<t)=P(X<t^2)$ So given the CDF of X you can just put $t^2$ in it. Basicly, For harder definitions given by $Y=g(X)$ where g is unnecessarily invertible I recommend drawing a plot of g, then when calculating $P(Y<t)$, for each value of $t$ look for the mathcing values of $X$ such that $g(X)<t$ and express $F_Y$ as a function of $F_X$.

Answer 3

About the so-called Third pattern:

Assuming the CDF and PDF of the random variable $X$ are $F_X$ and $f_X$ respectively, and that $X\geqslant0$ almost surely, the CDF $F_Y$ of $Y=\sqrt{X}$ is, by definition, such that, $F_Y(y)=0$ for every $y\lt0$ and, for every $y\geqslant0$, $$ F_Y(y)=P[Y\leqslant y]=P[X\leqslant y^2]=F_X(y^2). $$ Differentiating both sides yields the PDF of $Y$ as the function $f_Y$ such that $f_Y(y)=0$ if $y\lt0$ and, for every $y\geqslant0$, $$ f_Y(y)=\frac{\mathrm d}{\mathrm dy}F_Y(y)=\frac{\mathrm d}{\mathrm dy}F_X(y^2)=2yf_X(y^2). $$