GATE 2024 DA Q.59: solution and answer

GATE 2024 · DAQ.59 · 2 marks · NAT

Consider a joint probability density function of two random variables $X$ and $Y$

f_{X, Y} (x, y) = {2 x y, 0, 0 < x < 2, 0 < y < x otherwise

Then, $E [Y ∣ X = 1.5]$ is ______.

See the question in the paper

Tap to open full size

Answer and solution

ANSWER

$E [Y ∣ X = 1.5] = 1$ . As printed, $2 x y$ on $0 < y < x < 2$ integrates to $4$ , not $1$ . The scaling constant cancels in the conditional density $f_{Y ∣ X} (y ∣ x) = 2 y / x^{2}$ , so the answer is $1$ whether or not the joint is normalised.

Official answer: marks to all candidatesSource: Official answer key, GATE 2024 DA (IISc Bengaluru)

The support is the triangle $0 < y < x < 2$ , so once $X = x$ is fixed, $Y$ ranges over $(0, x)$ .
Check the total probability first. $\int_{0}^{2} \int_{0}^{x} 2 x y d y d x = \int_{0}^{2} x \cdot x^{2} d x = \int_{0}^{2} x^{3} d x = \frac{2 ^{4}}{4} = 4.$ The total is $4$ , not $1$ , so the printed function is not a valid joint density as it stands.
The natural reading is that the density is proportional to $2 x y$ on the triangle. Dividing by $4$ gives the valid joint density $f_{X, Y} (x, y) = \frac{x y}{2}$ for $0 < y < x < 2$ .
Marginal of $X$ : $f_{X} (x) = \int_{0}^{x} \frac{x y}{2} d y = \frac{x}{2} \cdot \frac{x ^{2}}{2} = \frac{x ^{3}}{4}, 0 < x < 2.$ Check: $\int_{0}^{2} \frac{x ^{3}}{4} d x = \frac{16}{16} = 1$ .
Conditional density of $Y$ given $X = x$ : $f_{Y ∣ X} (y ∣ x) = \frac{f _{X, Y} ( x , y )}{f _{X} ( x )} = \frac{x y /2}{x ^{3} /4} = \frac{2 y}{x ^{2}}, 0 < y < x .$ It integrates to $1$ : $\int_{0}^{x} \frac{2 y}{x ^{2}} d y = \frac{x ^{2}}{x ^{2}} = 1$ . The scaling constant cancels in this ratio, so the conditional density is the same whether or not the joint is normalised. That makes the answer well defined despite the misprint.
Conditional expectation: $E [Y ∣ X = x] = \int_{0}^{x} y \cdot \frac{2 y}{x ^{2}} d y = \frac{2}{x ^{2}} \cdot \frac{x ^{3}}{3} = \frac{2 x}{3} .$
Substitute $x = 1.5$ : $E [Y ∣ X = 1.5] = \frac{2 \times 1.5}{3} = 1.$

KEY IDEA

Given $X = x$ , the conditional density is proportional to the joint density with $x$ held fixed. On $0 < y < x$ that is proportional to $y$ , so $f_{Y ∣ X} (y ∣ x) = 2 y / x^{2}$ and $E [Y ∣ X = x] = 2 x /3$ . Any constant in front of the joint density cancels in this ratio.

WATCH OUT FOR

One mistake is taking $Y$ from $0$ to $2$ instead of $0$ to $x$ . Another is using the marginal mean $E [Y]$ instead of the conditional one. A third is to assume, without checking, that $2 x y$ is a valid density: it integrates to $4$ on this triangle, so $x^{3}$ is not the true marginal of $X$ (the valid one is $x^{3} /4$ ). A fourth is to read the conditional as uniform on $(0, 1.5)$ , which would give $0.75$ .

From the GATE 2024 DA paper. Every question of this paper · Download the full paper · All 8 Random variables questions, with notes

GATE 2024 DA Q.59: solution and answer

Answer and solution

More Random variables questions