Random Variables MCQ Quiz in मल्याळम - Objective Question with Answer for Random Variables - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

The correct answer is 6:5

 Key Points

• Odds are a ratio that express the likelihood of an event happening or not happening.
• The odds in favor of an event are the ratio of the number of ways the event can happen to the number of ways it cannot happen.
• The odds against an event are the opposite of the odds in favor, obtained by flipping the ratio.
• Odds can be converted to probabilities by dividing the number of favorable outcomes by the total number of outcomes.
• Odds are commonly used in betting, gambling, and statistical probability.
• Understanding odds can help in making informed decisions, evaluating risks, and estimating the likelihood of events.

Additional Information

• Odds are a ratio that express the likelihood of an event happening or not happening.
• The odds in favour of an event are the ratio of the number of ways the event can happen to the number of ways it cannot happen.
• If the odds in favor of an event A are 5:6, this means that for every 5 ways the event can happen
  • there are 6 ways it cannot happen.
• To find the odds against the event, we simply flip the ratio: Odds against the event = 6:5
• So, the odds against event A are 6:5.

Concept:

For a uniformly distributed random variable,

$E[X]=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x{f}_x(x)dx$

Calculation:

$E[\sin x]=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}sinx{f}_x(x)dx$

$={\displaystyle \frac{1}{2\pi }}\underset{-\pi }{\overset{\pi }{\int }}\sin xdx$

$={\displaystyle \frac{-1}{2\pi }}[\cos x{]}_{-\pi }^{\pi }$

${\displaystyle \frac{-1}{2\pi }}[\cos \pi -\cos (-\pi )]$

= 0

Calculation:

Clearly Y = X², hence X and Y are not independent.

(statement 3 is correct)

$E\left\{Z\right\}={\int }_{-2}^{2}f\left(Z\right)dz$

$={\int }_{-2}^2\frac{1}{4}dz$

E{Z} = 0

E{X} = E{Z} = 0

E{Y} = E{Z²}

$E\left\{Y\right\}={\int }_{-2}^2\frac{1}{4}{Z}^{2}dz$

= $\frac{1}{4}{\left(\frac{Z^3}{3}\right)}_{-2}^2$

= $\frac{1}{12}\left(8+8\right)=\frac{16}{12}$

= $\frac{4}{3}$

The covariance μ is:

μ  = E{(X – m_x)(Y – m_y)}

= E{(X – 0)(Y – 4/3)}

= $E\left\{\left(X\right)\left(Y-\frac{4}{3}\right)\right\}$

= $E\left\{XY-\frac{4X}{3}\right\}$

= $E\left\{Z^3-\frac{4Z}{3}\right\}$

= ${\int }_{-2}^2\frac{1}{4}\left(Z^3-\frac{4Z}{3}\right)dz$

= $\frac{1}{4}{\left[\frac{Z^4}{4}-\frac{4Z^2}{6}\right]}_{-2}^2$

$\frac{1}{4}\left[\left(\frac{16}{4}-\frac{16}{6}\right)-\left(\frac{16}{4}-\frac{16}{6}\right)\right]$

= 0

∴ as μ  = 0, ∴ the variances are uncorrelated.

Y = X²

Since the probability density function f_X(x) of random variable X has even symmetry.

So, we have:

$E\left[X^n\right]=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^n{f}_X\left(x\right)dx=0$      ---(1)

Where n is an odd integer.

Therefore, we have the mean of random variable X as

X̅ = E[X] = 0

Also, the mean value of random variable Y is given by

Y̅ = E[Y] = E[X²] = X̅²

$\overline{Y}={\overline{X}}^2+{\sigma }_X^2={\sigma }_X^2$

Now, we check the orthogonality and correlation for the given random variables.

Orthogonal:

Two random variables X and Y are orthogonal if E[XY] = 0

For the given problem, we have:

E[XY] = E[X X²] = E[X³]       ---(2)

Substituting n = 3 in equation (1), we get

E[X³] = 0

So, substituting this value in equation (2), we obtain

E[XY] = 0

Therefore, the variables X and Y are orthogonal

Correlation:

Two random variables X and Y are uncorrelated only if their correlation coefficient is zero; i.e.

$\rho =\frac{cov\left[X,Y\right]}{{\sigma }_X{\sigma }_Y}=0$

Now, for the given variables, we obtain the covariance as

cov[X, Y] = E[(X – X̅)(Y – Y̅)]

$=E\left[\left(X-0\right)\left(X^2-{\sigma }_X^2\right)\right]$

$=E\left[X\left(X^2-{\sigma }_X^2\right)\right]$

$=E\left[X^3\right]-{\sigma }_X^{2}E\left[X\right]$

$=0-{\sigma }_X^2\times 0=0$

Thus, the correlation coefficient ρ = 0

Therefore, the random variables are uncorrelated.

Concept:

The relation between CDF [cumulative Distribution OR Density function] and PDF.

$$\begin{gathered} F_X\left(x\right)={\int }_{-\mathrm{\infty }}^n{f}_X\left(x\right)\cdot dx \\ \{\begin{array}{c}{F}_X\left(x\right)=CDF\\ F_X\left(x\right)=PDF\end{array} \end{gathered}$$

CDF is a Non-decreasing function

F_x(x) = P[X ≤ x]

F_x(∞) = 1

Calculation:

We know, for valid PDF :

${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{F}_X\left(x\right)=1$ 

$F_X\left(x\right)=1\left[1-\frac{\left|x\right|}{1}\right]=1-\left|x\right|$  

∴ $F_X\left(x\right)={\int }_{-\mathrm{\infty }}^x\left[1-\left|x\right|\right]\cdot dx$ 

Observation:

a) Since F_x(∞) = 1 & CDF is Non-decreasing function, the close option 3 and 4 neglected.

b) Since integration of straight line is curve, therefore option (1) will be answer. 

$\begin{array}{rl}& \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\mathrm{p}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}=1,Nowp\left(x\right)=\frac{A}{8}\left(x+3\right)\\ & \therefore \frac{1}{8}\underset{-3}{\overset{5}{\int }}\left(Ax+3A\right)dx=1or\frac{A}{8}{\left[\frac{x^2}{2}+3x\right]}_{-3}^5=1\\ & Or,\frac{A}{8}[12.5+15-4.5+9]=1\\ & \therefore A=\frac{1}{4}\\ & E\left[x\right]=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\mathrm{x}\mathrm{p}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}=\underset{-3}{\overset{5}{\int }}\frac{x}{32}\left(x+3\right)dx=\frac{1}{32}\left[\frac{x^3}{3}+\frac{3x^2}{2}\right]\\ & =\frac{1}{32}\left[\frac{125}{3}+\frac{75}{2}+\frac{27}{3}-\frac{27}{2}\right]=2.333\\ & Var\left[x\right]={{\sigma }_x}^2=E\left[x^2\right]-E{\left[x\right]}^2=\underset{-3}{\overset{5}{\int }}{x}^2\mathrm{p}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}-{\left(\frac{7}{3}\right)}^2=\underset{-3}{\overset{5}{\int }}\frac{x^2}{32}\left(x+3\right)\mathrm{d}\mathrm{x}-\frac{49}{9}\\ & {{\sigma }_x}^2=\frac{1}{32}{\left[\frac{x^4}{4}+x^3\right]}_{-3}^5-\frac{49}{9}=\frac{1}{32}\left[\frac{625}{4}+125-\frac{81}{4}+27\right]-\frac{49}{9}\\ & =3.556\end{array}$

Concept: n^th moment of a random variation is

$E\left[X^n\right]=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^n{f}_X\left(x\right)dx.$

Now for Gaussian distribution with zero mean

$\mathrm{E}\left[{\mathrm{X}}^{\mathrm{n}}\right]=\left\{\begin{array}{c}0\mathrm{i}\mathrm{f}\mathrm{n}\mathrm{i}\mathrm{s}\mathrm{o}\mathrm{d}\mathrm{d}.\\ 1.3.5\dots \left(\mathrm{n}-1\right){\sigma }^2\mathrm{i}\mathrm{f}\mathrm{n}\mathrm{i}\mathrm{s}\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}\end{array}\right\}$

Calculation: For 4^th moment n = 4. therefore n-1 will be equals to 3.

$E\left[X^4\right]=1.3.{\sigma }^2$

= 3.4

= 12

hence 4th moment is 12

Given the function,

$f_X\left(x\right)=\frac{k}{x^2}u\left(x-k\right)$

Now, we check the validity of this function

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{f}_X\left(x\right)dx$

$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{k}{x^2}\left(x-k\right)dx$

$=\underset{k}{\overset{\mathrm{\infty }}{\int }}\frac{k}{x^2}dx$

$=k{\left[\frac{x^{-1}}{-1}\right]}_k^{\mathrm{\infty }}=\frac{k}{k}$

= 1

Concept:

The probability of a function can be calculated either by integrating its pdf (Probabilty density function) or directly form the CDF

Plotting the given function in the given range.

Application:

Clearly the above picture is of CDF to get pdf, differentiate CDF

${\mathrm{f}}_{\mathrm{x}}\left(\mathrm{x}\right)=\frac{1}{\sqrt{\pi }}{\mathrm{e}}^{\left(-{\mathrm{x}}^2+\mathrm{x}-\mathrm{a}\right)}$

${\mathrm{f}}_{\mathrm{x}}\left(\mathrm{x}\right)$ can be rewritten as:

${\mathrm{f}}_{\mathrm{x}}\left(\mathrm{x}\right)=\left[\frac{1}{\sqrt{\pi }}{\mathrm{e}}^{-{\left(\mathrm{x}-\frac{1}{2}\right)}^2}\right]{\mathrm{e}}^{-\left(\mathrm{a}-\frac{1}{4}\right)}$

Now, for fX(x) to be a valid pdf;

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\mathrm{f}}_{\mathrm{x}}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}={\mathrm{e}}^{-\left(\mathrm{a}-\frac{1}{4}\right)}\times \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{1}{\sqrt{\pi }}{\mathrm{e}}^{-{\left(\mathrm{x}-\frac{1}{2}\right)}^2}\mathrm{d}\mathrm{x}=1$

Now, $\frac{1}{\sqrt{\pi }}{\mathrm{e}}^{-{\left(\mathrm{x}-\frac{1}{2}\right)}^2}$ can be interpreted as a normal distribution with mean 0.5 and variance 0.5, i.e, N(0.5, 0.5).

Thus, substituting $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{1}{\sqrt{\pi }}{\mathrm{e}}^{-{\left(\mathrm{x}-\frac{1}{2}\right)}^2}\mathrm{d}\mathrm{x}=1$

We have,

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\mathrm{f}}_{\mathrm{x}}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}={\mathrm{e}}^{-\left(\mathrm{a}-\frac{1}{4}\right)}=1$

a = 0.25