Considering all the symbols with their usual meanings, match the following : Codes :List - I List - II (a) \(f_{X}(x)\) (i) \(\displaystyle\int_{-\infty}^{\infty}f_{X,Y}(x,y)\,dx\) (b) \(\displaystyle\int_{-\infty}^{\infty}f_{X}(x)\,dx\) (ii) \(\displaystyle\int_{-\infty}^{a}f_{X}(x)\,dx\) (c) \(F_{X}(a)\) (iii) \(\displaystyle\int_{-\infty}^{\infty}f_{X,Y}(x,y)\,dy\) (d) \(f_{Y}(y)\) (iv) 1
(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
Two ideas generate all four answers: marginalisation and the total-probability normalisation.
(a) \(f_{X}(x)\) is the marginal density of X. To recover it from the joint density you integrate the other variable out — here y, leaving a function of x:
\(f_{X}(x)=\int_{-\infty}^{\infty}f_{X,Y}(x,y)\,dy\) → (iii)
(d) By symmetry, the marginal of Y integrates x out:
\(f_{Y}(y)=\int_{-\infty}^{\infty}f_{X,Y}(x,y)\,dx\) → (i)
The variable of integration is the one you are eliminating, not the one you keep — this is the single point the question is testing, and it is why (a) and (d) are so easy to interchange.
(b) Integrating a density over its entire range must give certainty:
\(\int_{-\infty}^{\infty}f_{X}(x)\,dx=1\) → (iv)
This is the normalisation condition that every valid pdf must satisfy, along with \(f_{X}(x)\ge0\).
(c) The cumulative distribution function accumulates the density up to the point of interest:
\(F_{X}(a)=P(X\le a)=\int_{-\infty}^{a}f_{X}(x)\,dx\) → (ii)
Note the finite upper limit — that is what distinguishes it from (b).
| Quantity | Obtained by | Code |
|---|---|---|
| Marginal fX(x) | Integrate out y | (iii) |
| Total area | Integrate over all x | (iv) |
| CDF FX(a) | Integrate up to a | (ii) |
| Marginal fY(y) | Integrate out x | (i) |
The order (iii), (iv), (ii), (i) is option 4.
The relations that tie the set together : the density is the derivative of the distribution, \(f_{X}(x)=\dfrac{dF_{X}(x)}{dx}\); the CDF rises monotonically from 0 at \(-\infty\) to 1 at \(+\infty\), the endpoint value being exactly statement (b); and the two marginals recover the joint density only when X and Y are independent, in which case \(f_{X,Y}(x,y)=f_{X}(x)f_{Y}(y)\). In general the joint density carries information about the dependence that neither marginal retains.
Hence, the correct match is (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i).
Consider the following probability density function for a random variable x.
(a) \(f_{1}(x)=1\ ;\ -1\le x\le 1\)
(b) \(f_{2}(x)=x\ ;\ 0\le x\le 1\)
(c) \(f_{3}(x)=(1-x)\ ;\ -1\le x\le 1\)
Arrange the above functions in terms of the increasing value of mean of random variable x.
The density function of a random variable is given by \(p(x)=Ke^{-\frac{x^{2}}{2}}\) for \(-\infty \lt x \lt \infty\). The value of K should be :
If the odds in favour of any random event A are 5 ∶ 6, then the odds against the event are:
If random variable X follows binomial distribution with parameter n and p with mean 15 and variance 10, then the value of mode is
Consider the following probability density function for a random variable x.
(a) \(f_{1}(x)=1\ ;\ -1\le x\le 1\)
(b) \(f_{2}(x)=x\ ;\ 0\le x\le 1\)
(c) \(f_{3}(x)=(1-x)\ ;\ -1\le x\le 1\)
Arrange the above functions in terms of the increasing value of mean of random variable x.
The density function of a random variable is given by \(p(x)=Ke^{-\frac{x^{2}}{2}}\) for \(-\infty \lt x \lt \infty\). The value of K should be :