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.
(c), (b), (a)
This question was cancelled by UGC. The official answer key records code 9 against it, meaning every option was credited and all candidates were awarded the two marks whatever they marked. The discussion below therefore gives the best-supported reading of the item rather than a uniquely correct option.
Compute the mean of each density from the definition \(\mu=\displaystyle\int x\,f(x)\,dx\).
(a) Uniform on [−1, 1]. The density is symmetric about the origin, so by inspection
\(\mu_{1}=\int_{-1}^{1}x\cdot 1\,dx=\left[\dfrac{x^{2}}{2}\right]_{-1}^{1}=0\)
(Strictly, a proper uniform density on this interval is \(\tfrac{1}{2}\) rather than 1, but the normalising constant cannot change a mean that is zero by symmetry.)
(b) Triangular on [0, 1]. Here the density rises linearly and the whole support is positive, so the mean must be positive:
\(\mu_{2}=\int_{0}^{1}x\cdot x\,dx=\left[\dfrac{x^{3}}{3}\right]_{0}^{1}=\dfrac{1}{3}\)
(c) Decreasing ramp on [−1, 1]. This one weights negative values more heavily than positive ones, so its mean must be negative:
\(\int_{-1}^{1}x(1-x)\,dx=\int_{-1}^{1}x\,dx-\int_{-1}^{1}x^{2}\,dx=0-\dfrac{2}{3}=-\dfrac{2}{3}\)
Since \(\int_{-1}^{1}(1-x)dx=2\), normalising gives \(\mu_{3}=-\dfrac{1}{3}\) — negative either way.
| Density | Mean | Sign |
|---|---|---|
| (c) 1 − x on [−1, 1] | −1/3 | Negative — smallest |
| (a) uniform on [−1, 1] | 0 | Zero — middle |
| (b) x on [0, 1] | +1/3 | Positive — largest |
The true increasing order is therefore (c), (a), (b) — and that sequence is not among the four options. The question as printed is defective. Of what is offered, option 3, (c), (b), (a), is the only code that correctly places the negative-mean density (c) first, and it is taken as the intended answer; the pairing is flagged for checking against the official key.
The reasoning worth carrying away, which needs no integration at all : a density supported symmetrically about zero and itself symmetric has zero mean; a density living entirely on the positive axis has a positive mean; and a density that tilts weight toward the negative side of a symmetric interval has a negative mean. Those three observations order the list on sight.
Hence, per the code set the answer is (c), (b), (a), the strictly correct order being (c), (a), (b).
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